Sponsor Area
24x < 100
∴ (i) When
∴ Solution set is
(ii) When
The solution set is
Solve -12x > 30, when
(i) x is a natural number. (ii) x is an integer.
-12x > 30
(∵ -12 < 0)
∴ (i) When , there is no solution
∴ The solution set is .
(ii) When
The solution set is
Solve 5x – 3 < 7, when
(i) x is an integer. (ii) x is a real number.
5x – 3 < 7
5x – 3 < 7 5x < 10
x<2
∴ (i) When
The solution set is
(ii) When
∴ The solution set is
Solve 3x + 8 > 2, when
(i) x is an integer. (ii) x is a real number.
3x + 8 > 2 3x > 2 - 8
3x > - 6
x > -2
∴ (i) When
x = -1, 0, -1, 2, 3, ........
The solution set is
(ii) When
The solution set is
3 (2 – x) ≥ 2 (1 – x)
(∵ -1 < 0)
Hence, the solution is
Solve the following inequality and show the graph of the solution on the number line.
3 (1 - x) < 2 (x + 4)
3 (1 - x) < 2 (x + 4)
(∵ -5 < 0)
x > - 1
Hence, the solution of the inequality is
Graph on the number line is:
Solve the following inequality and show the graph of the solution on the number line.
Solve the following inequality and represent the solution graphically on the number line.
Sponsor Area
Solve the following inequality and represent the solution graphically on the number line.
5 (2x – 7) – 3 (2x + 3) ≤ 0 , .
5 (2x – 7) – 3 (2x + 3) ≤ 0
...(i)
Also,
...(ii)
From (i) and (ii), we have
∴ The solution set is [-7, 11]. Graphically, we have
Solve the following pair of inequality and represent the solution graphically on the number line:
Solve the following pair of inequality and represent the solution graphically on the number line:
Let the two consecutive odd positive integers be x and x + 2
Since both the integers are smaller than 10
∴ x < 10 and x + 2 < 10 x < 10 and x < 8
x < 8 ...(i)
Also, ...(ii)
From (i) and (ii), we have
But x is odd. x = 5, 7 and x + 2 = 7, 9
Hence, the pair of odd numbers are satisfying the given conditions are 5, 7 and 7, 9.
Let the consecutive even positive integers be x and x + 2
...(i)
Their sum is less than 20. x + (x + 20) < 20
2x + 2 < 20
2x < 18
x < 9 ...(ii)
From (i) and (ii), we have 5 < x < 9,
But, x is given, ∴ x = 6, 8
When x = 6, x + 2 = 8
When x = 8, x + 2 = 10
Hence, the two pairs that satisfy the given conditions are: 6, 8 and 10.
Let marks obtained by the student in annual examination be x.
∴ average of marks = But, minimum average marks = 60
Hence, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.
Let the shortest side of the triangle be of length x cm The length of the longest side = 3x cm
The length of the third side = (3x - 2) cm
It is given that the perimeter is at least 61 cm Perimeter
Hence, the minimum length of the shortest side = 9 cm.
Let the length of the first piece (or the shortest piece) be x cm Length of the second piece = (x + 3) cm
Also, the length of the third piece = 2x cm. The maximum length of the board available = 91 cm.
...(i)
Also, the third piece is to be at least 5 cm longer than the second.
...(ii)
From (i) and (ii), we have
Hence, the shortest board (or the first board) can have a minimum length of 8 cm and a maximum length of 22 cm.
Quantity of 45% acid solution = 1125 litres
Quantity of acid =
Suppose x litres of water is added
∴ Quantity of acid =
Total quantity of mixture = (1125 + x) litres
% of acid content =
According to the problem,
Now, ...(i)
Also,
...(ii)
From (i) and (ii), we have
Hence, a minimum of 562.5 litres and a maximum of 900 litres of water be added so as to move a solution with 25% to 30% of acid content.
We have
It is given that 200 < T < 300
or 200 < 30 + 25(x - 3) < 300
or 170 < 25 (x - 3) < 270
or
or 6.8 + 3 < x < 10.8 + 3
or 9.8 < x < 13.8
Hence, the depth will be between 9.8 km and 13.8 km.
We have ...(i)
Now, CA = 12 and (given)
Hence, MA is at least 9.6 but not more than 16.8.
It is given that
Putting we get
or
or
or 86 < F < 95
Thus, the required range of temperature is between
Let Sunita's marks in the fith examination be x out of 100 marks. Sunita's marks in the first four examinations are 87, 92, 94 and 95
Average marks obtained by Sunita =
To receive grade 'A' in a course Sunita has to get an average of 90 marks or more
∴
Hence, Sunita must obtain a minimum of 82 marks in the fifth examination to get grade 'A' in the course.
Solution not provided.
Ans. 7, 9.
Solution not provided.
Ans. 6,8; 8,10; 10,12
Solution not provided.
Ans. 87
Solution not provided.
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Solution not provided.
Ans. 600
Solution not provided.
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Solution not provided.
Ans. Between 7 km and 9.5 km
Solution not provided.
Ans. More than 120 litres but less than 300 litres.
Sponsor Area
Solve the following inequality:
4x + 3 < 6x + 7, For (i) (ii)
(iii)
Solution not provided.
Ans. (i) (ii)
(iii)
Solve the following inequality:
For (i) (ii)
(iii)
Solution not provided.
Ans. (i) {8, 9, 10,.........} (ii) {8, 9, 10, .....} (iii)
Solve the following inequality and represent them on the number line:
3x - 7 > x + 3
Solution not provided.
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Solve the following inequality and represent them on the number line:
4x - 7 < 3 - x
Solution not provided.
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Solve the following inequality and represent them on the number line:
Solution not provided.
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Solve the following inequality and represent them on the number line:
Solution not provided.
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Solve the following inequality and represent them on the number line:
Solution not provided.
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Solve the following inequality and represent them on the number line:
2 (2x + 3) - 10 < 6 (x - 2)
Solution not provided.
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Solve the following inequality and show the solution graphically on the number line:
x + 3 >0; 2x < 14
Solution not provided.
Ans. (-3, 7)
Solve the following inequality and show the solution graphically on the number line:
Solution not provided.
Ans. [4, 9]
Solve the following inequality and show the solution graphically on the number line:
Solution not provided.
Ans. (4, 12]
Solve the following inequality and show the solution graphically on the number line.
Solution not provided.
Solve the following inequality and show the solution graphically on the number line.
Solution not provided.
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Solve the following inequality and show the solution graphically on the number line.
Solution not provided.
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Solve the following inequality and show the solution graphically on the number line.
Solution not provided.
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Solve the following inequality and show the solution graphically on the number line:
Solution not provided.
Ans. [2, 3]
Solve the following inequality and show the solution graphically on the number line:
Solution not provided.
Ans. (5, 10]
Solve the following inequality for (i) (ii)
(iii)
and show that the graphs of the solutions on the number line. : 3x - 2 < 2x + 1
3x - 2 < 2x + 1 3x - 2x < 1 + 2
x < 3
(i) Solution for
The solution set is The graph on the number line is:
(ii) x = 2, 1, 0, -1, -2, -3, ............
∴ The solution set is The graph on the number line is:
(iii) The graph on the number line is:
Solve the following inequality for (i) (ii)
(iii)
and show that the
graphs of the solutions on the number line. :
Centre (h, k) (-2, 3)
h = -2, k = 3, Radius, r = 4
∴ The equation of circle is :
or
Centre (h, k)
∴ Equation of circle is :
or
or
or
or
Hence, the equation of circle is:
Centre (h, k) (-a, -b)
h = -a, k = -b
Diameter of circle =
∴ Equation of circle is :
or
Hence, the equation of circle is:
The centre of circle is C (h, k) (2, 2)
h = 2, k =2
Since the circle passes through the point A (4, 5).
∴ Radius of the circle is : r = CA
∴ Equation of circle is:
or
Hence, the equation of circle is
Let the equation of the circle be ...(i)
Since it passes through (0, 0)
∴ ...(ii)
Since it passes through (4, 0)
∴ ...(iii)
But, from (ii),
∴ From (iii), we have
Since it passes through (0, 3)
∴
∴ From (ii), we have
or
Hence, the equation of circle is:
or
or
The circle passes through the origin and makes intercepts a with x-axis and b with y-axis. OA = a, OB = b
∴ The co-ordinates of A are (a, 0) and of B are (0, b)
Now, we have to find the equation of a circle which passes through the points O (0, 0), A (a, 0), B (0, b).
Let the equation of the circle
...(i)
Since it passes through (0, 0)
∴
...(ii)
Since it passes through A (a, 0)
∴
...(iii)
But, from (ii),
∴ from (iii), we have
Since the circle passes through (0, b)
∴
∴ from (ii), we have
∴ The equation of circle is:
or
Hence, the equation of circle is :
The co-ordinates of the centre of the circle are (3, -2) h = 3, k = -2 ...(i)
Let r be the radius of the circle
Area of the circle = (Given)
∴ Radius, r = 7 ...(ii)
Equation of the circle in the standard form is:
or
which is the required equation of the circle.
Radius, r of the circle = 5 ...(i)
The centre is the point of intersection of straight lines
2x - y - 5=0 and 3x + 2y = 4
Solving these two equations simultaneously, we have
4x - 2y = 10
3x + 2y = 4
Adding, we get 7x = 14 or x = 2 4 - y - 5 = 0 or y = -1
∴ The co-ordinates of the centre are (2, -1) h = 2, k = -1 and r = 5
The equation of the circle in the standard form is:
or
which is the required equation of the circle.
Let AB and EF be the two diameters of the circle and have equations
x - 2y = 7 and x + 2y = -1 respectively.
The centre C of the circle is the point of intersection of the two diameters
∴ For C,
x - 2y = 7 ...(i)
x + 2y = -1 ...(ii)
Adding, we get 2x = 6 x = 3
and 3 - 2y = 7 y = -2
∴ The co-ordinates of centre C of the circle are (3, -2)
Also, the circle passes through point P(1, 4)
∴ Radius of the circle, r = CP = h = 3, k = -2,
The equation of the circle is:
or
which is the required equation of the circle.
Centre of the required circle is C (-2, 3)
Let line AB with equation 3x - 4y - 2 = 0 touches the circle at point L
∴ p = r
where p, is length of perpendicular CL and r is the radius
The equation of circle is:
or
Hence, the equation of circle is:
Let C be the centre of the circle
∴ C is the mid-point of AB.
∴ Co-ordinates of C, (h, k)
(h, k)
(1, -3)
∴ h = 1, k = -3
Also, radius, r = CB
∴ The equation of circle is:
or
Hence, the equation of the circle is :
Let the equation of the circle be ...(i)
It passes through (2, 3)
...(ii)
It passes through (-1, 1)
...(iii)
From (ii) and (iii), we have
...(iv)
Also, centre (h, k) lies on the line x - 3y -11 = 0 h - 3k - 11 = 0 ...(v)
Multiplying (v) by 6, we get
6h - 18k - 66 =0 ...(vi)
Subtracting (vi) from (iv), we get
∴ from (v),
Also, from (ii),
Hence, the equation of circle is:
or or
or
Hence, the equation of circle is:
Let the equation of the circle be ...(i)
The circle passes through A (5, 7)
...(ii)
The circle passes through B (6, 6)
...(iii)
The circle passes through point C (2, -2)
...(iv)
Equating (ii) and (iii), we have -10h - 14k + 74 = -12h - 12k + 72 2h - 2k + 2 = 0
h - k + 1 = 0 ...(v)
Equating (iii) and (iv), we have
-12h - 12k + 72 = -4h + 4k + 8 8h + 16k - 64 = 0
h + 2k - 8 =0 ...(vi)
Subtracting (v) from (vi), we get
3k - 9 = 0 k = 3
Also, from (vi), h + 2 (3) - 8 = 0 h = 2
So, the centre of the circle is :
It passes through point A (5, 7) r = distance between the centre and the point.
r =
Hence, the equation of the circle is:
or
or
The given equation of the circle is:
Comparing it with
∴ the centre of circle is
The radius of circle is
The given equation of circle is :
Comparing it with the standard equation we have
∴ The centre of circle is and radius is 6.
The given equation of the circle is:
or
Comparing it with the standard equation of circle: we have
Hence, the centre of circle is and the radius is 7.
The given equation of the circle is:
Comparing it with the standard equation of circle, we get
Hence, the co-ordinates of the centre are
The given equation of the circle is: (Note this step)
Comparing it with the standard equation of the circle, we have
∴ the centre of the circle is
Sponsor Area
The given equation is : ...(i)
It can be written as
Completing the squares in each bracket, we have
or
or
Comparing it with the standard form we get
Hence, the centre of the circle is and its radius is 3a.
Does the point (-2.5, 3.5) lie inside, outside or on the circle ?
The equation of circle is
Comparing it with the standard equation of circle, we have
h = 0, k=0 and r = 5
∴ Centre of circle is O (0, 0) radius = 5
Distance of point A (-2.5, 3.5) or from the centre is
Distance of point from centre is less than the radius.
Hence, the point lies inside the circle.
Does the point (2, -1) lie inside, or on the circle ?
The equation of the circle is:
Comparing it with the standard form of circle , we have
∴ The centre is C (-3, 1) and radius is r = 4
Distance of point A (2, -1) from the centre.
∴ Distance of point from the centre is greater than the radius.
Hence, the point lies outside the circle.
Find the equation of the circle whose centre and radii are:
Centre (-3, -2), radius 7
Solution not provided.
Ans.
Find the equation of the circle whose centre and radii are:
Centre (a, a), radius
Solution not provided.
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Find the equation of the circle whose centre and radii are:
Centre
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Find the equation of circle which has 3x - 2y - 4 = 0 and 2x + y - 5 = 0 as two of its diameters and a radius of
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Ans. (a, a) ; 2a
Solution not provided.
Ans. (2, 4) ;
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Solution not provided.
Ans. (0, 0) ; r
Solution not provided.
Ans. outside, d>r.
Solution not provided.
Ans. on, d=r.
Solution not provided.
Ans. inside, d<r
Let line l be the directrix with equation
3x + 4y + 1 = 0 and S (2, 5) be the focus.
Take a point on the parabola
∴ By definition, PS = 'PM d = p
Hence, the locus of P or the equation of parabola is :
The vertex is at (0, 0) and the focus is at (-3, 0). For both the points y = 0
∴ Axis of the parabola is y = 0 or the x-axis. Equation of the parabola in the standard form is
Also, focus
Hence, the equation of the parabola is
The vertex of the parabola is at O(0, 0) and the focus at S (0, -2). For both the points, x = 0. The x-axis of the parabola is along x = 0 or the y-axis.
The equation of the parabola in the standard form is
...(i)
Also, the focus
Hence, the equation of the parabola is:
The vertex of the parabola is O (0, 0) and the axis is along x-axis (y = 0). The equation of the parabola is
...(i)
Since it passes through the point (2, 3)
∴
Hence, from (i), the equation of the parabola is :
Sponsor Area
The vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The axis of the parabola is y-axis (x = 0) The equation of the parabola is
...(i)
Since it passes through the points (2, -3)
∴
Hence, from (i), the equation of the parabola is:
The vertex is (0, 0) and the focus on positive x-axis is S (a, 0), a>0
∴ Equation of parabola is ...(i)
Latus rectum = 4a = (∵ a > 0)
Hence, from (i), the equation of parabola is:
The vertex is at (0, 0) and the focus is on negative y-axis is S(0, -a), a>0
∴ The equation of the parabola is ...(i)
Latus rectum = 4a = 11
Hence, from (i), the equation of parabola is:
The focus is S(6, 0) which lies on x-axis.
The directrix is x + 6 = 0 which is a line parallel to y-axis i.e., perpendicular to x-axis.
∴ The parabola is of the standard form
Also, focus (a, 0) (6, 0) directrix x + a = 0 is x + 6 = 0
a = 6
Hence, the equation of the parabola is
Alternative method:
Let line l be the directrix with equation x + 6 = 0
The focus is S (6, 0). Take a point on the parabola.
From P, draw PM perpendicular on directrix l and join PS. By definition of parabola, PS = PM d = p
Squaring both sides, we get
Hence, the locus of P i.e., the equation of parabola is
The focus of parabola is (0, -3) which lies on y-axis. Directrix of the parabola is y - 3 = 0 which is parallel to x-axis.
∴ The equation of the parabola is of the standard form ...(i)
Focus is (0, -a) (0, -3) and directrix y - a = 0 is y - 3 = 0
a = 3
Hence, form (i), the equation of the parabola is
Alternative method:
Let l be the directrix with equation y - 3 = 0.
S (0, -3) is the focus.
Take a point on the parabola. From P, draw PM perpendicular on the directrix l and join PS. By definition of parabola, PS = PM
Hence, the equation of locus of P i.e, equation of parabola is
Let the point be
∴ [∵ Ordinate is 3 times the abscissa]
lies on the parabola
When
Hence, the coordinates of the point are (2, 6).
The equation of the parabola is 4a = 12 or a = 3
Focus S is (0, a) (0, 3)
OS = 3
LL' = length of the latus rectum
= 4a = 12
Hence, area of
An equilateral triangle is inscribed in the parabola , where one vertex is at the vertex of the parabola. Find (a) the length of the side of the triangle, (b) area of triangle ABC.
Let OAB be the triangle with side p.
In
Now,
and
∴ Point A is
Since it lies on
∴
Hence, (i) each side of , (ii) area of
A parabolic reflection with diameter PR = 20 cm and OQ = 5 cm is shown below:
Here, vertex of the parabola is (0, 0)
Let the focus be S (a, 0)
Let the equation of the parabola be
Now, PR = 20 cm PQ = 10 cm
Also, OQ = 5 cm
∴ Point P is (5, 10)
Since it lies on the parabola
∴
∴ Focus is S (5, 0) which is same as point Q.
Equation of parabola is y2 = 4ax. The focus is at S (a, 0).
∴ OS = a
Let coordinates of point L be and of L' be
with
∴
∴ Coordinates of L are (a, 2a) and of L' are (a, -2a).
Equation of parabola is 4a = 8
a = 2
Focus is at point (0, -2).
OS = -2
Let the co-ordinates of L be and of L' be
Since L lies on
∴ Coordinates of L are (4, -2) and of L' are (-4, -2)
Let AB be the parabolic arch having O at the vertex and the vertical line OY as the axis.
The parabola open upwards
∴ Its equation is of the form ...(i)
Width of the arch, LM = 5 m
OM = 2.5 m
Height of the arch, BM = 10 m
∴ Co-ordinates of point B are (2.5, 10)
Since point B lies on the parabola
∴
∴ From (i), the equation of the parabola is:
or ...(ii)
We have to find the width PQ of the arch at a distance ON = 2 m from the vertex.
Let PQ = d NQ =
∴ Co-ordinates of point Q are
Putting it in (ii), we get
Hence, the width of the arch = d = or 2.23 (approx).
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If a parabolic reflector is in diameter and 20 cm deep, find the distance of its focus S from the vertex.
Solution not provided.
Ans. 6 cm
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Ans. 8 square units.
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Ans. 512 square units.
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus - (-3, 0)
Axis - (y = 0)
Tangent at vertex - (x = 0)
directrix - (x - 3 = 0)
latus rectum = 12
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus - (0, 9)
Axis - (x = 0)
Tangent at vertex - (y = 0)
directrix - (y + 9 = 0)
latus rectum - (36)
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus -
Axis - (x = 0)
Tangent at vertex - (y = 0)
directrix -
latus rectum - (9)
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus -
Axis - (y = 0)
Tangent at vertex - (x = 0)
directrix -
latus rectum -
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus -
Axis - (x = 0)
Tangent at vertex - (y = 0)
directrix -
latus rectum - (9a)
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus - (2, 0)
Axis - (y = 0)
Tangent at vertex - (x = 0)
directrix - (x + 2 = 0)
latus rectum - (8)
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus -
Axis - (x = 0)
Tangent at vertex - (y = 0)
directrix -
latus rectum -
Find the coordinates of vertex, focus, equations of axis, tangent at vertex, directrix, latus rectum of the following equation:
Solution not provided.
Ans. Vertex - (0, 0)
Focus - (-1, 0)
Axis - (y = 0)
Tangent at vertex - (x = 0)
directrix - (x - 1 = 0)
latus rectum - (4)
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The cable of a uniformly loaded suspension bridge hang in the form of a parabola.
The roadway which is horizontal and 100 m long, is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to a roadway 18 m from the middle.
Let AOB be the cable of a uniformly loaded suspension bridge as shown in the figure.
Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the road way as shown in the figure.
The longest supporting wire = AL = BM = 30 m
The shortest supporting cable = OC = 6 m
Length of the roadway = LM = 100 m
∴ LC = CM = 50 m
Let O be the vertex and OY be the axis of the parabola
∴ Equation of the parabola is ...(i)
ON = CM = 50
BN = BM - MN = BM - OC = 30 - 6 = 24
∴ Co-ordinates of point B are (50, 24)
Since it lies on the parabola (i)
∴
∴ From (i), the equation of the parabola is:
or ...(ii)
Let the length of the supporting wire PQ at a distance of 18 m be h.
∴ OR = CQ = 18 m
PR = PQ - RQ = PQ - OC = h - 6 Co-ordinates of point P are (18, h - 6)
Since it lies on parabola (ii)
∴
1944 + 3750 = 625 h
Let point P be
Givent that
Squaring both sides, we get
Squaring again, we get
Hence, the locus of point P is
which is the equation of an ellipse in the standard form
Equation of ellipse is
∴ Equation of ellipse is
Now,
Co-ordinates of foci
Focus,
Focus,
Co-ordinates of vertices
Length of major axis = 2a = 2 (6) = 12
Length of minor axis = 2b = 2 x 4 = 8
Let the eccentricity be e. Now, c = ae
∴
Length of latus rectum,
Equation of ellipse is
∴ Equation of ellipse in the standard form is
Co-ordinates of foci
Focus,
Focus,
Co-ordinates of vertices
Vertex,
Vertex,
Length of major axis = 2a = 2 x 10 = 20
Length of minor axis = 2b = 2 x 5 = 10
Let e be the eccentricity. Now, c = ae
∴
Latus rectum,
The equation of ellipse is :
Now,
∴ Equation of ellipse in standard form is
Focus,
Focus,
Vertex,
Vertex,
Length of major axis = 2a = 2(4) = 8
Length of minor axis = 2b = 2(1) = 2
Let the eccentricity be e.
Now, c = ae
Length of latus rectum =
Equation of ellipse is :
Now,
∴ Equation of ellipse in the standard form is
(i) Focus,
Focus,
(ii) Vertex,
Vertex,
(iii) Length of major axis = 2a = 2(3) = 6
Length of minor axis = 2b = 2(2) = 4
(iv) Let the eccentricity be e.
Now, c = ae,
∴
(v) Length of latus rectum
Find: (i) the co-ordinates of foci.
(ii) the co-ordinates of vertices.
(iii) the length of major axis and minor axis.
(iv) the eccentricity.
(v) latus rectum of ellipse.
Equation of ellipse is :
Now,
The equation of the ellipse in standard form is
Now,
(i) Focus,
Focus,
(ii) Vertex,
Vertex,
(iii) Length of major axis = 2a =
Length of minor axis = 2b =
(iv) Let the eccentricity be e
Now, c = ae
(v) Length of latus rectum =
Find: (i) the co-ordinates of foci.
(ii) the co-ordinates of vertices.
(iii) the length of major axis and minor axis.
(iv) the eccentricity.
(v) latus rectum of ellipse.
Equation of ellipse is:
Now,
Also,
Equation of the ellipse in standard form is
(i) Focus,
Focus,
(ii) Vertex,
(iii) Length of major axis = 2a =
Length of minor axis = 2b =
(iv) Let the eccentricity be e
Now,
(v) Length of latus rectum =
The foci (± 4, 0) lie on x-axis. Equation of ellipse in standard form is
...(i)
Foci:
Vertices:
Now,
∴ From (i), equation of ellipse is
The foci (0, ± 5) lie on y-axis.
∴ Equation of ellipse in standard form is ...(i)
Foci:
Vertices:
Now,
Hence, from (i), the equation of the ellipse is
Ends of major axis lie on x-axis.
∴ Equation of ellipse in standard form is ...(i)
Ends of major axis (vertices) are
Ends of minor axis are
Also,
∴ from (i), equation of ellipse is ...(ii)
The foci are
Ends of major axis lie on y-axis.
∴ Equation of ellipse in the standard form is
Ends of major axis are
Ends of minor axis are
Hence, from (i), the equation of ellipse is
Also,
∴ The foci are
The foci (± 5, 0) are on x-axis.
∴ The equation of ellipse in standard form is ...(i)
Foci are
Length of major axis 2a = 26 a = 13 or
Also,
∴ From (i), equation of ellipse is :
Let e be the eccentricity
∴ c = ae 5 = 13e
The foci (0, ± 6) lie on y-axis.
∴ The equation of ellipse in standard form is ...(i)
The Foci are
Length of minor axis 2b = 16 b = 8 or
Also,
∴ From (i), equation of ellipse is :
Let e be the eccentricity
∴ c = ae 6 = 10e
The major axis of the ellipse is along the x-axis.
∴ The equation of ellipse is ...(i)
It passes through (6, 3)
...(ii)
It passes through (12, 0)
∴ From (ii),
Hence, the equation of ellipse, from (i), is:
The major axis is along the y-axis.
∴ The equation of ellipse in the standard form is ...(i)
It passes through (2, 2)
...(ii)
It passes through (1, 4)
...(iii)
Multiplying both sides of (ii) by 4, we get
...(iv)
Subtracting (iv) from (iii), we get
∴ From (iii),
Hence, from (i), the equation of ellipse is:
The foci lie on x-axis.
∴ The equation of ellipse in standard form is
Foci
Also, a = 4 (given)
Using,
Hence, from (i), the equation of ellipse is :
Also, c = ae 3 = 4e
Equation of ellipse in the standard form is [∵ foci lie on x-axis]
Here, c = 4, b = 3
∴
Hence, from (i), equation of ellipse is :
Let the eccentricity of the ellipse be e.
∴ c = ae 4 = 5e
Let P be any point on the ellipse with the two foci at
and
Verify that
Let point P be (x, y). Focus is (-c, 0),
is (c, 0)
∴
Now,
But,
∴
∴
Hence, , which verifies the ellipse property,
For ellipse verify
where
and
be the foci of the ellipse and P is any point on it.
Let P(x, y) be any point on the ellipse ...(i)
...(ii)
Focus is (-c, 0), Focus
is (c, 0)
From (i),
Now,
∴ Focus Focus
∴
Hence,
Let be the two points where the flag posts are fixed on the ground.
Take the line joining points and
as the x-axis, the mid-point of line segment
as the origin O and the line through O, perpendicular to
as y-axis.
∴ Co-ordinates of points are (-4, 0) and (4, 0) respectively.
Take any point on the track.
∴
Squaring both sides, we get
Squaring both sides, we get
Hence, the equation of the locus of is :
which is the equation of the ellipse in standard form
Width of the elliptical arch = 8 m AA' = 8m
2a = 8m
a = 4
Height at the centre = 2m OB = 2m
b = 2
∴ the equation of ellipse is ...(i)
Let the height at P be h.
∴ Coordinates of L are (2.5, h)
It lies on ellipse (i)
Hence,
In the following, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Solution not provided.
Ans. (i) (ii)
(iii) 8, 6 (iv)
(v)
In the following, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Solution not provided.
Ans. (i) (ii)
(iii) 8, 6 (iv)
(v)
In the following, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Solution not provided.
Ans. (i) (ii)
(iii) 6, 2 (iv)
(v)
In the following, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Solution not provided.
Ans. (i) (ii)
(iii) 6, 2 (iv)
(v)
In the following, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Solution not provided.
Ans. (i) (ii)
(iii) 10, 4 (iv)
(v)
In the following, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Solution not provided.
Ans. (i) (ii)
(iii) 8, 6 (iv)
(v)
Solution not provided.
Ans. (i) (ii)
(iii) 40, 20 (iv)
(v) 10
Tips: -
Solution not provided.
Ans. (i) (ii)
(iii) 14, 12 (iv)
(v)
Solution not provided.
Ans. (i) (ii)
(iii) 20, 10 (iv)
(v) 5
Solution not provided.
Ans. (i) (ii)
(iii) 26, 10 (iv)
(v)
Find the equation for the ellipse that satisfies the given condition:
Vertices , foci
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Vertices , foci
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Ends of major axis ends of minor axis
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Ends of major axis ends of minor axis
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Ends of major axis 18, foci
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Length of minor axis 8, foci
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Foci,
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Foci
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
b = 5, c = 3, axis along the y-axis.
Solution not provided.
Ans.
Find the equation for the ellipse that satisfies the given condition:
Major axis on x-axis, passing through points A (4, 3) and B (6, 2).
Solution not provided.
Ans.
Find the eccentricity of an ellipse if its latus rectum is of its minor axis.
Solution not provided.
Ans.
Find the equation of an ellipse whose foci are centre is at origin and which passes through the point (4, 1).
Solution not provided.
Ans.
P is any point on the ellipse Verify that
where
and
are the foci of the ellipse.
Solution not provided.
Let the point be .
The fixed points are A (4, 0), B (-4, 0). The given condition is :
PA - PB = 2
Squaring both sides, we get
Squaring again, we get
Hence, the locus of point P is
which is the standard form of hyperbola in the form .
Let the point be It is given that
Squaring both sides, we get
Squaring again, we get
Hence, the locus of is
which is the equation of a hyperbola in the standard form
The equation of hyperbola is :
which is of the form
∴ The foci and vertices lie on x-axis.
So,
∴
The co-ordinates of foci are:
The co-ordinates of the vertices are:
The length of transverse axis = 2a = 2 x 6 = 12
The length of conjugate axis = 2b = 2 x 8 = 16
Let the eccentricity be e, we know that c = ae 10 = 6e
The length of latus rectum
The equation of hyperbola is :
which is the equation of hyperbola in the standard form
∴ The foci and vertices lie on y-axis.
So,
Also,
The co-ordinates of foci are:
The co-ordinates of vetices are:
The length of transverse axis = 2a = 2 x 4 = 8
The length of conjugate axis = 2b = 2 x 7 = 14
Let the eccentricity be e
∴ c = ae
Length of latus rectum =
Since the foci and the vertices
lie on y-axis.
∴ The equation of hyperbola in the standard form is ...(i)
Comparing foci:
Comparing vertices:
Using we have
∴ From (i), the equation fo hyperbola is
Since foci and the vertices
lie on the x-axis.
∴ The equation of the hyperbola is ...(i)
Comparing foci :
Comparing vertices :
Now,
Hence, from (i), the equation of hyperbola is
The foci lie on x-axis.
∴ The equation of the hyperbola in standard form is ...(i)
The foci are
The distance between the two foci = 2c = 10 (given) c = 5
Also, the length of transverse axis = 2a = 8 a = 4
Now,
Hence, from (i), the equation of hyperbola is :
Find the equation of a hyperbola whose vertices lie on y-axis, centre is at the origin, the distance between the foci is 16 and eccentricity is
The vertices of the hyperbola lie on the y-axis.
∴ The equation of the hyperbola in standard form is ...(i)
The foci are
The distance between them is 2c = 16 (given) c = 8
Eccentricity, e =
∴ c = 8 ae = 8
or
Now,
Hence, from (i), the equation of the hyperbola is: or
Find the co-ordinates of foci, the vertices, the length of transverse axis, the length of conjugate axis, the eccentricity, the latus rectum of the hyperbola:
The hyperbola is : which is of the form
∴ Its foci and vertices lie on x-axis.
The foci are :
The vertices are:
The length of transverse axis = 2a = 2 x 4 = 8
The length of conjugate axis = 2b = 2 x 3 = 6
Let the eccentricity be e. We know that c = ae 5 = 4e
Length of latus rectum =
Find the co-ordinates of foci, the vertices, the length of transverse axis, the length of conjugate axis, the eccentricity, the latus rectum of the hyperbola:
The equation of hyperbola is which is of the form
∴ The foci and vertices lie on the y-axis.
Also,
The co-ordinates of foci are :
The co-ordinates of vertices are:
The length of transverse axis = 2a = 2 x 3 = 6
The length of conjugate axis = 2b =
Let the eccentricity be e. We know that c = ae 6 = 3(e)
e = 2
Length of latus rectum =
The foci lie on the x-axis.
∴ The standard equation of the hyperbola is of the form : ...(i)
The foci are
The latus rectum ...(ii)
Now, (∵
)
(∵ a = -9 is not possible)
∴
Also,
Hence, from (i), the equation of the hyperbola is:
The foci are of the hyperbola lie on the y-axis.
∴ The standard equation of the hyperbola is: ...(i)
The foci are
Also, the latus rectum is
...(ii)
Now, (∵
)
(∵ a = -8 is not possible)
Also,
Hence, from (i), the equation of the hyperbola is or
.
The vertices of the hyperbola lie on x-axis.
∴ The standard equation of the hyperbola is ...(i)
The vertices are
Also, eccentricity
∴ c = ae =
Now,
Hence, from (i), the equation of hyperbola is : .
The vertices of the hyperbola lie on the y-axis.
∴ The standard equation of the hyperbola is ...(i)
Vertices are
Also, eccentricity
Now,
Also,
Hence, from (i), the equation of the hyperbola is:
or
Find the equation of a hyperbola whose foci are and which pass through point (2, 3).
The foci of the hyperbola are which lie on y-axis.
∴ The standard equation of the hyperbola is ...(i)
It passes through (2, 3)
...(ii)
The foci are
But, ...(iii)
Using (iii) in (ii), we have
When from (iii),
(not possible)
When
∴
Hence, from (i), the equation of the hyperbola is: or
For the hyperbola 3X2 – 2y2 = 1.
Find: (i) the lengths of the transverse and conjugate axes.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum.
Equation of the hyperbola is
which is of the standard form
Also,
(i) Length of transverse axis
Length of conjugate axis = 2b =
(ii) The co-ordinates of foci are:
(iii) The co-ordinates of vertices are:
(iv) Let the eccentricity be e.
∴
(v) Length of latus rectum =
P(x, y) is a point on hyperbola Verify that
where
are the foci of the hyperbola.
Solution not provided.
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola .
Solution not provided.
Ans. (i) 6, 8 (ii) (iii)
(iv)
(v)
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola .
Solution not provided.
Ans. (i) 8, 2 (ii) (iii)
(iv)
(v)
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola.
Solution not provided.
Ans. (i) 4, 6 (ii) (iii)
(iv)
(v) 9
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola.
Solution not provided.
Ans. (i) 8, 6 (ii) (iii)
(iv)
(v)
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola.
Solution not provided.
Ans. (i) (ii)
(iii)
(iv)
(v) 2
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola.
Solution not provided.
Ans. (i)
(ii)
(iii)
(iv)
(v)
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola.
Solution not provided.
Ans. (i) 8, 12
(ii)
(iii)
(iv)
(v) 18
Find:
(i) the lengths of transverse axis and conjugate axis.
(ii) the co-ordinates of foci.
(iii) the co-ordinates of vertices.
(iv) the eccentricity.
(v) the length of latus rectum of the following hyperbola.
Solution not provided.
Ans. (i) 10, 6 (ii) , (iii)
(iv)
(v)
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans.
Find the equation of a hyperbola whose latus rectum is 8 and eccentricity is
Solution not provided.
Ans.
The foci of a hyperbola coincide with the foci of ellipse Find the equation of the hyperbola if its eccentricity is 2.
Solution not provided.
Ans.
Solution not provided.
Ans. (6, -5)
Solution not provided.
Ans.
Find (a) the co-ordinates of the centre (b) the radius (c) the equation of the circle drawn on the line segment joining points A (– 1, 2) and B (3, – 4) as diameter.
Solution not provided.
Ans. (a) (1, -1) (b) (c)
Solution not provided.
Ans.
Solution not provided.
Ans.
Find (i) the co-ordinates of the focus.
(ii) the equation of directrix
(iii) the latus rectum for the parabola
Solution not provided.
Ans. (i) (0, -4) ; (ii) y - 4 = 0; (iii) 16
Find (i) the co-ordinates of the focus.
(ii) the equation of directrix.
(iii) the latus rectum for the parabola.
Solution not provided.
Ans.(i) (-2, 0)
(ii) x - 2 =0
(iii) 8
Find (i) the co-ordinates of the focus.
(ii) the equation of directrix.
(iii) the latus rectum for the parabola.
Solution not provided.
Ans. (i)
(ii)
(iii)
Find (i) the co-ordinates of the focus.
(ii) the equation of directrix.
(iii) the latus rectum for the parabola.
Solution not provided.
Ans. (i)
(ii)
(iii) 6
Check whether the points (2, 2), (4, 4) and lie inside, outside or on the cirlcle
The given equation of the circle is
The centreo of the circle is (0, 0) and radius of circle is 3
For point (2, 2).
Distance of point (2, 2) from centre (0, 0) = (radius of circle)
∴ Point (2, 2) lies inside the circle
For point (4, 4),
Distance of point (4, 4) from centre (0, 0) = (radius of circle)
∴ Point (4, 4) lies outside the circle
For point
Distance of point from the centre of the circle (0, 0).
(radius of the circle).
∴ Point lies on the circle.
Mid-point of AB is
∴ Centre C of the circle is
Radius of the circle = CA
Hence, the equation of circle is :
The centre of the circle is (2, -3)
The circle touches the straight line 3x - 4y - 3 = 0
Hence, the equation of the circle is
2x + 6 > 0 2x > -6
x > - 3
The graph of x = -3, is obtained by drawing a vertical line at x = -3 on the x-axis and dotted line is drawn.
(0, 0) substituted in the inequality gives 0 > -3 which is true.
∴ The region is the one that contains (0, 0)
(or alternatively the inequality is x > -3, we shall mark the region towards the right of the vertical line)
The graph is:
3y + 6 < 0 3y < - 6 or y < - 2
The line drawn will be at y = -2 on y-axis and it will be horizontal.
The inequality is y<-2, so dotted line is drawn O (0, 0) substituted in y > -2 0 < -2 which is false.
∴ The region represented is the one which does not contain (0, 0).
(or alternatively y < -2 is towards the lower side of the horizontal line)
The graph of inequality is :
First, we draw the graph of straight line 2x + 3y = 0
Now, let us test a point B(1, 1) which is not true.
From these obseravtions, the graph is given below in which the shaded region represents the solution of given inequality which does not contain point B(1, 1).
First, we draw the graph of straight line 4x - 3y = 0
x 0 3
y 0 4
Now, let us take a point B(1, 1)
∴ 4(1) - 3(1) 0 or 1
0 which is true.
From these observations, the required solution is represented by the shaded portion of the graph given below which contains point B(1, 1).
First, we draw the dotted line to represent the graph of x + y = 5
x 0 5
y 5 0
Now, let us take a point O(0, 0)
Substitute in given inequality, 0 + 0 < 5, i.e., 0 < 5 which is true.
Therefore, origin O is a solution of given inequality.
From these observations, the graph is given below, in which the shaded portion represents the solution of the given inequality which contains O(0, 0).
First, we draw the dark line to represent the graph of x - y = 2
x 0 2
y -2 0
Now, let us take a point O(0, 0)
Substitute in given inequality which is not true.
Therefore, origin is not the solution of given inequality.
From these observations, the graph is given below, in which the shaded portion represents the solution of the given inequality.
Solve the following inequality graphically in two-dimensional plane:
– 3x + 2y ≥ – 6
First, we draw the dark line to represent the graph of straight line -3x + 2y = -6
x 0 2
y -3 0
Now, let us take a point O(0, 0), i.e., origin.
Substitute in given inequality, -3(0) + 2(0) ≥ -6, i.e., 0 ≥ -6 which is true. Therefore, origin O(0, 0) is solution of given inequality. From these observations, the graph is given below, in which the shaded region represents the solution of the given inequality.
Solve the following inequality graphically in two-dimensional plane:
3y – 5x < 30
First, we draw the dotted line to represent the graph of straight line 3y - 5x = 30
x 0 -6
y 10 0
Now, let us take a point O(0, 0) i.e, origin.
Substitute in given inequality, 3(0) - 5(0) < 30 or 0 < 30 which is true. Therefore, origin O(0, 0) is the solution of given inequality. From these observations, the graph is given ahead in which shaded region represents the solution of the given inequality.
Solve the following inequality graphically in two-dimensional plane:
2x – 3y > 6
First, we draw a dotted line to represent graph of line 2x - 3y = 6.
x 0 3
y -2 0
Now, let us take a point O(0, 0). Substitute in given inequality. 2(0) - 3(0) > 6 or 0 > 6 which is not true. Therefore, origin is not the solution of given inequality. From these observations, the graph is given below in which the shaded region represents the solution of the given inequality.
Solve the following inequality graphically in two-dimensional plane:
y + 8 ≥ 2x
First, we draw a dark line to represent the graph of straight line y + 8 = 2x.
i.e., 2x - y = 8
x 0 4
y -8 0
Now, let us take a point O(0, 0), i.e, origin.
Substitute in given inequality, 0 + 8 ≥ 2(0) or 8 ≥ 0 which is true.
Therefore, origin is the solution of given inequality.
From these observations, the graph is given below, in which shaded region represents the solution of the given inequality.
The given linear inequalities are :
x ≥ 2 ...(i)
y ≤ 3 ...(ii)
First, we draw the graph of straight lines x = 2 and y = 3.
For x ≥ 2, we mark the region towards right of line x = 2.
For y ≤ 3, we mark the region below the line y = 3.
Now, we shade the common region.
Thus, the graphical solution is given below:
The given linear inequalities are:
x ≥ 3 ...(i)
y ≥ 2 ...(ii)
First, we draw the lines x = 3 and y = 2.
For x ≥ 3, we mark the region on right side of line x = 3.
For y ≥ 2, we mark the region above the line y = 2.
Now, we shade the common region.
Thus, the graphical solution is:
First, we draw the straight line 3x + 2y = 12
x 0 4
y 6 0
Now, let us take the point O(0, 0), i.e., origin.
Substituting O(0, 0) in inequality
3x + 2y ≤ 12, we get
3(0) + 2(0) ≤ 12 or 0 ≤ 12 which is true.
∴ origin is solution of inequality
3x + 2y ≤ 12.
Therefore, we mark that region which contains region.
For x ≥ 1, we mark the region on right side of line x = 1.
For y ≥ 2, we mark the region above the line y = 2.
Now, we shade the common region.
Thus, the graphical solution is the triangular region ABC as shown in the figure.
First, we draw the line 4x + 5y = -20
x 0 -5
y -4 0
Now, consider the point O(0, 0), i.e, origin.
Substitute in inequality 4x + 5y ≥ -20,
i.e, 4(0) + 5(0) ≥ - 20 or 0 ≥ - 20 which is true.
∴ origin is solution of inequality
4x + 5y ≥ -20
Therefore, we mark that region which contains origin.
For x + 1 ≤ 0 or x ≤ -1, we mark the region on left side of line x = -1.
For y ≤ -1, we mark the region below the line y = -1.
Now, we shade the common region. The required solution is given by the triangular region ABC as shown below in the graph.
First, we draw the graph of line x + y = 9
x 0 9
y 9 0
Now, consider the point O(0, 0), i.e, origin.
Substitute in inequality x + y ≤ 9, i.e., 0 + 0 ≤ 9 or 0 ≤ 9 which is true. Therefore, origin is solution of inequality x + y ≤ 9. Therefore, we mark that region which contains origin.
Now, we draw the line y = x or y - x = 0
x 0 1 2
y 0 1 2
Let us consider the point (1, 0).
Substitute it in inequality y ≥ x, i.e., 0 ≥ 1 which is not true.
∴ We mark that region which does not contain point (1, 0).
For x ≥ 0, we mark the region on right side of the line x = 0.
Now, we shade the common region. Thus, the solution is given by the triangular region OAB as shown in the graph.
Clearly, x ≥ 0, y ≥ 0 means that the solution set lies in the first quadrant.
Now, we draw the graph of line 3x + 4y = 60
x 0 20
y 15 0
Now, we consider the point O(0, 0) i.e., origin.
Substitute in inequality 3x + 4y ≤ 60, i.e., 3(0) + 4(0) ≤ 60 or 0 ≤ 60 which is true.
Therefore, we mark that region which contains origin.
Now, we draw the graph of line x + 3y = 30
x 0 30
y 10 0
Now, we consider the point O(0, 0), i.e, origin.
Substitute in inequality x + 3y ≤ 30, 0 + 3(0) ≤ 30 or 0 ≤ 30 which is true. Therefore, we mark that region which contains origin.
Now, we shade the common region. The solution is given by the shaded portion OABC of the graph given below;
First, we draw the graph of line x - 2y = 3
x 0 3
y -3/2 0
Now, consider a point O(0, 0), i.e, origin.
Substitute in given inequality x - 2y ≤ 3, i.e, 0 - 2(0) ≤ 3 or 0 ≤ 3 which is true. Therefore, we mark that region which contains origin. Now, we draw the graph of line 3x + 4y = 12.
x 0 4
y 3 0
Now, we consider a point O(0, 0), i.e., origin.
Substitute in given inequality 3x + 4y ≤ 12, i.e., 3(0) + 4(0) ≤ 12 or 0 ≤ 12 which is true.
∴ we mark that region which contains region.
For x ≥ 0, we mark the region on right side of line x = 0.
For y ≥ 1, we mark the region above the line y = 1
Now, we shade the common region.
The solution is given by the triangular region ABC as shown in the graph.
First, we draw the graph of line 3x + 2y = 150
x 0 50
y 75 0
Now, we consider a point O(0, 0), i.e, origin.
Substitute it in inequality 3x + 2y ≤ 150, i.e., 3(0) + 2(0) ≤ 150 or 0 ≤ 150 which is true.
∴ we mark that region which contains origin.
We draw the graph of line x + 4y = 80
x 0 80
y 20 0
Now, we consider a point O(0, 0), i.e, origin.
Substituting it in inequality x + 4y ≤ 80, we get
0 + 4(0) ≤ 80 or 0 ≤ 80 which is true.
∴ we mark that region which contains origin.
For x ≥ 15, we mark the region on right side of the line x = 15
For y ≥ 0, we mark the region above the line y = 0
Now, we shade the common region.
The required solution is given by shaded region ABCD as shown in the graph.
We draw the graph of line 2x + y = 6
x 0 3
y 6 0
Now, we consider a point O(0, 0), i.e, origin.
Substitute in inequality 2x + y ≥ 6, i.e. 7, 2(0) + 0 ≥ 6 or 0 ≥ 6 which is not true.
∴ we mark that region which does not contain origin.
Now, we draw the graph of line 3x + 4y = 12
x 0 4
y 3 0
Now, we consider a point O(0, 0), i.e., origin.
Substituting in given inequality, 3x + 4y ≤ 12, we get 3(0) + 4(0) ≤ 12 which is true.
Therefore, we mark that region which contains origin. Now, we shade the common region. Thus, the required solution set is given by the shaded portion of graph given below.
We draw the dotted line to represent the graph of 2x - y = 1
x 0 1/2
y -1 0
Now, we consider a point O(0, 0), i.e., origin.
Substitue it in inequality 2x - y > 1, we get 2 (0) - 0 > 1 or 0 > 1 which is not true.
Therefore, we mark tht region, which does not contain origin.
We draw the dotted line to represent the graph of line x - 2y = -1
x 0 -1
y 1/2 0
Now, we consider a point O(0, 0) i.e., origin.
Substitute it in inequality x - 2y < - 1,
i.e., 0 - 2(0) < -1 or 0 < -1 which is not true.
∴ we mark that region which does not contain origin.
Now, we shade the common region. Thus, the solution is given by the shaded region of the graph given below;
Consider the line 2x + y = 2. Clearly, the shaded portion and the origin are on the opposite sides of the graph. So, we have its inequality as:
[∵ 2(0) + 0 ≥ 2 is not true]
Now, consider the line x + 2y = 8.
Origin and shaded region are on the same side of the line x + 2y = 8.
So, its inequality is:
[∵ 0 + 2(0) ≤ 8 is true]
Finally, consider the line x - y = 1
Origin and shaded region are on the same side.
So, its inequality is:
[∵ 0 - 0 ≤ 1 is true]
Also, x ≥ 0 and y ≥ 0
Thus, the linear inequalities are:
2x + y ≥ 2, x + 2y ≤ 8, x - y ≤ 1, x ≥ 0 and y ≥ 0
Solving the following inequalities. Also represent the solutions on the number line.
3x - 2 > x + 5
Solution not provided.
Ans.
Solving the following inequalities. Also represent the solutions on the number line.
-4x + 1 ≤ 2 (1 - x)
Solution not provided.
Ans.
Solving the following inequalities. Also represent the solutions on the number line.
Solution not provided.
Ans. ,
Solving the following inequality. Also represent the solutions on the number line.
Solution not provided.
Ans.
Solve x + 1 < 8 - x, where
(i) (ii)
(iii)
Solution not provided.
Ans. (i) {1, 2, 3} (ii) {........ -3, -2, -1, 0, 1, 2, 3} (iii)
Solve the following system of simultaneous linear inequality and represent its solution on the number line.
2x + 3 < x - 1 and x + 4 > 2x - 3
Solution not provided.
Ans. (i)
Solve the following system of simultaneous linear inequality and represent its solution on the number line.
-3x + 2 ≥ 0 and 2 - x < 0
Solution not provided.
Ans. No solution
Make the region represented by the following inequality:
2x - y ≥ 1
Solution not provided.
Ans.
Make the region represented by the following inequality:
2x + 3y ≤ 6
Solution not provided.
Ans.
Exhibit graphically, the solution set of the linear inequalities.
4x + 3y ≤ 12, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Solution not provided.
Ans.
Solve the following system of linear inequalities graphically.
x + y ≤ 5, 4x + y ≥ 4, x + 5y ≥ 5, x ≤ 4, y ≤ 3
Solution not provided.
Ans.
Find the linear inequalities for which the shaded area in the diagram is the solution set.
Solution not provided.
Ans. 2x + 3y ≥ 6, x + 2y ≤ 6, -3x + 2y ≤ 3, x - 2y ≤ 2, x ≥ 0, y ≥ 0
Find all pairs of consecutive odd positive integers both of which are smaller than 10, such that their sum is greater than 11.
Solution not provided.
Ans. 5 and 7, 7 and 9
Draw the graph of the following inequalities: 2x + 3y - 6 ≤ 0 and x + y ≥ 1
Solution not provided.
Ans.
Draw the graph of the following inequalities : x + y ≤ 2, 2x + y ≥ 1, x - y ≤ 0, x ≥ 0 and y ≥ 0.
Solution not provided.
Ans.
Find the linear inequalities for which the shaded area in the figure given below is the solution.
3x + 2y ≤ 6, 2x + 3y ≤ 6, x ≥ 0, y ≥ 0
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
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Solution not provided.
Ans.
Find the linear inequalities for which the shaded area in the figure given below is a solution.
4x + 9y ≤ 36, x + 3y ≥ 6, 3x + y ≥ 6, 4x - 3y ≤ 12
The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2 =6 on any tangent to it is
(x2-y2)2 = 6x2+2y2
(x2-y2)2 = 6x2 -2y2
(x2+y2)2 = 6x2+2y2
(x2+y2)2 = 6x2-2y2
C.
(x2+y2)2 = 6x2+2y2
Equation of ellipse is,
The slope of the perpendicular drawn from the centre (0,0) to (h,k) is k/h.
Since, both the lines are perpendicular,
The slope of the line touching both the parabolas y2 = 4x and x2-32y is
1/2
3/2
1/8
2/3
A.
1/2
For parabola, y2 = 4x
Let y = mx + 1/m be tangent line and it touches the parabola x2=-32y
The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point
(-5,2)
(2,-5)
(5,-2)
(-2,5)
C.
(5,-2)
(x − 3)2+ y2+ λy = 0
The circle passes through (1, − 2)
⇒ 4 + 4 − 2λ = 0 ⇒ λ = 4
(x − 3)2+ y2+ 4y = 0
⇒ Clearly (5, − 2) satisfies.
Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 − Tn = 10, then the value of n is
7
5
10
8
B.
5
Given,Tn = nC3
Tn+1 = n+1C3
therefore,
Tn+1-Tn = n+1C3-nC3 =10
⇒nC2 + nC3-nC3 = 10
[∵nCr +nCr-1 =n+1Cr]
nC2 =10
nC2 =5C2
n =5
Given A circle, 2x2 + 2y2= 5 and parabola,
Statement I An equation of a common tangent to these curves is
Statement II If the line is the common tangent, then m satisfies m4-3m2+2 =0
Statement I is true, Statement II is true; Statement II is a correct explanation for statement I
Statement I is true, Statement II is true; Statement II is not a correct explanation for statement I
Statement I is true, Statement II is false
Statement I is false, Statement II is true
B.
Statement I is true, Statement II is true; Statement II is not a correct explanation for statement I
Equation of circle can be rewritten as,
x2 +y2 = 5/2
Let common tangent be
The perpendicular from centre to the tangent is equal to radius,
therefore,
thus, both Statements are correct.
Statement I An equation of a common tangent to the parabola and the ellipse 2x2 +y2 =4 is
.
Statement II If the line is a common tangent to the parabola
and the ellipse 2x2 +y2 =4, then m satisfies m4 +2m2 =24
Statement 1 is false, statement 2 is true
Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
Statement 1 is true, statement 2 is false
C.
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
If z ≠ 1 and is real, then the point represented by the complex number z lies
either on the real axis or on a circle passing through the origin
on a circle with centre at the origin
either on the real axis or on a circle not passing through the origin
on the imaginary axis
A.
either on the real axis or on a circle passing through the origin
Let z = x + iy (∴ x ≠ 1 as z ≠ 1)
z2= (x2 – y2) + i(2xy)
its imaginary part = 0 is real
⇒ 2xy (x – 1) – y(x2– y2) = 0
⇒ y(x2+ y2– 2x) = 0
⇒y 0; x2+ y2 – 2x = 0
∴ z lies either on real axis or on a circle through origin.
The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is
10/3
3/5
6/5
5/3
A.
10/3
An ellipse is drawn by taking a diameter of the circle (x–1)2 + y2 = 1 as its semiminor axis and a diameter of the circle x2 + (y – 2)2 = 4 as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is
4x2+ y2 = 4
x2 +4y2 =8
4x2 +y2 =8
x2+4y2 =16
D.
x2+4y2 =16
The equation of the ellipse.
Diameter of circle (x-1)2 +y2 = 1 is 2 and that of circle x2 +(y-2)2 = 4 is 4 units.
⇒ Semi-minor axis of ellipse, b =2 units are semi major axis of ellipse, a 4 unite
Hence, the equation of the ellipse is
For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is
there is a regular polygon with r/R = 1/2
there is a regular polygon with
there is a regular polygon with r/R = 2/3
there is a regular polygon with
C.
there is a regular polygon with r/R = 2/3
hyperbola passes through the point P(√2, √3) and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point
C.
Equation of hyperbola is
foci is (±2, 0) hence ae = 2, ⇒ a2e2 = 4
b2 = a2(e2 – 1)
∴ a2 + b2 = 4
Hyperbola passes through √2,√3
Let a, b, c ∈ R. If f(x) = ax2 + bx + c is such
that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, ∀x,y ∈ R,then is equal to
225
330
165
190
B.
330
f(x) = ax2 + bx + c
f(1) = a + b + c = 3
Now f(x + y) = f(x) + f(y) + xy
put y = 1
f(x + 1) = f(x) + f(1) + x
f(x + 1) = f(x) + x + 3
Now
f(2) = 7
f(3) = 12
Now
Sn = 3 + 7 + 12 + ..... tn ...(1)
Sn = 3 + 7 + ......tn – 1 + tn ...(2)
On subtracting (2) from (1)
tn = 3 + 4 + 5 + ....... upto n terms
The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directives is x= –4, then the equation of the normal to it at (1,3/2) is
x + 2y = 4
2y – x = 2
4x – 2y = 1
4x + 2y =7
C.
4x – 2y = 1
Eccentricity of ellipse =1/2
Now, a/e = -4
⇒ a = 4 x (1/2) = 2
therefore, b2 = a2(1-e2)
= a2 (1-(1/4)) = 3
Equation of normal at (1, 3/2)
y-3/2 = 2(x – 1)
⇒ 2y – 3 = 4x – 4
⇒ 4x – 2y = 1
Statement −1 is false, Statement −2 is true
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
Statement − 1 is true, Statement − 2 is false.
B.
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
The sum of the series 20C0 – 20C1 + 20C2 – 20C3 + …… - ….. + 20C10 is-
– 20C10
0
B.
(1 + x)20 = 20C0 + 20C1x + … + 20C10x10 + …+ 20C20x20
put x = − 1,
0 = 20C0 − 20C1 + … − 20C9 + 20C10 − 20C11 + … + 20C20
0 = 2 (20C0 − 20C1 + … − 20C9) + 20C10
⇒ 20C0 − 20C1 + … + 20C10 =
If the expansion in powers of x of the function is a0 + a1x + a2x2 + a3x3 + … , then an is
D.
(1-ax)-1(1-bx)-1 = (1+ax+a2x2+.....)(1+bx+b2x2+....)
therefore coefficient of xn = bn +abn-1 +a2bn-2 +.....+an-1b +an =
The value of ,where [x] denotes the greatest integer not exceeding x is
af(a) − {f(1) + f(2) + … + f([a])}
[a] f(a) − {f(1) + f(2) + … + f([a])}
[a] f([a]) − {f(1) + f(2) + … + f(a)}
af([a]) − {f(1) + f(2) + … + f(a)}
B.
[a] f(a) − {f(1) + f(2) + … + f([a])}
Let a = k + h, where [a] = k and 0 ≤ h < 1
{f(2) − f(1)} + 2{f(3) − f(2)} + 3{f(4) − f(3)}+…….+ (k−1) – {f(k) − f(k − 1)} + k{f(k + h) − f(k)}
= − f(1) − f(2) − f(3)……. − f(k) + k f(k + h)
= [a] f(a) − {f(1) + f(2) + f(3) + …. + f([a])}
The coefficient of the middle term in the binomial expansion in powers of x of (1 +αx)4 and of (1−αx )6 is the same if α equals
-5/3
3/5
-3/10
10/3
C.
-3/10
The coefficient of x in the middle term of expansion of (1+ αx)4= 4C2.α2
The coefficient of x in middle term of the expansion of 3(1− αx) = 6C3 (−α)3
According to question
Sponsor Area
Sponsor Area