Sponsor Area
A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2. Then the length of the semi−major axis is
8/3
2/3
5/3
4/3
A.
8/3
Major axis is along x-axis.
Consider a family of circles which are passing through the point (-1, 1) and are tangent to x-axis. If (h, K) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interva
0 < k < 1/2
k ≥ 1/2
– 1/2 ≤ k ≤ 1/2
k ≤ ½
B.
k ≥ 1/2
Equation of circle (x − h)2+ (y − k)2 = k2
It is passing through (− 1, 1) then
(− 1 − h)2+ (1 − k)2= k2
h2+ 2h − 2k + 2 = 0
D ≥ 0
2k − 1 ≥ 0 ⇒ k ≥ 1/2.
The differential equation of all circles passing through the origin and having their centres on the x-axis is
C.
General equation of all such circles is
x2+ y2 + 2gx = 0.
Differentiating, we get
For the Hyperbola 
which of the following remains constant when α varies?
Eccentricity
Directrix
Abscissae of vertices
Abscissae of foci
D.
Abscissae of foci
a2 = cos2α and b2 = sin2α
coordinates of focii are (± ae, 0)
∴ b2= a2(e2– 1) ⇒ e = secα.
Hence abscissae of foci remain constant when α varies.
The equation of a tangent to the parabola y2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is
(−1, 1)
(0, 2)
(2, 4)
(−2, 0)
D.
(−2, 0)
Point must be on the directrix of the parabola. Hence the point is (−2, 0)
In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is
3/5
1/5
2/5
4/5
A.
3/5
2ae = 6 ⇒ ae = 3
2b = 8 ⇒ b = 4
b2= a2(1 − e2)
16 = a2 − a2e2
a2= 16 + 9 = 25
a = 5
∴e = 3/a = 3/5
If the lines 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 are two diameters of a circle of area 49π square units, the equation of the circle is
x2 + y2 + 2x − 2y − 47 = 0
x2 + y2 + 2x − 2y − 62 = 0
x2 + y2 − 2x + 2y − 62 = 0
x2 + y2 − 2x + 2y − 47 = 0
D.
x2 + y2 − 2x + 2y − 47 = 0
Point of intersection of 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 is (1 , − 1), which is the centre of the circle and radius = 7.
∴ Equation is (x − 1)2 + (y + 1)2 = 49
⇒ x2 + y2 − 2x + 2y − 47 = 0.
Area of the greatest rectangle that can be inscribed in the ellipse 
2ab
ab
a/b
A.
2ab
Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is
y2 – 4x + 2 = 0
y2 + 4x + 2 = 0
x2 + 4y + 2 =
x2 – 4y + 2 = 0
A.
y2 – 4x + 2 = 0
P = (1, 0) Q = (h, k) such that
k2 = 8h
Let (α, β) be the midpoint of PQ
an ellipse
a circle
a straight line
a parabola
C.
a straight line
As given 
⇒ distance of z from origin and point (0,1/3) is same hence z lies on the bisector of the line joining points (0, 0) and (0, 1/3). Hence z lies on a straight line.
The locus of a point P (α, β) moving under the condition that the line y = αx + β is a tangent to the hyperbola 
an ellipse
a circle
a parabola
a hyperbola
D.
a hyperbola
Tangent to the hyperbola 
Given that y = αx + β is the tangent of hyperbola
⇒ m = α and a2 m2 – b2 = β2
∴ a2 α2 – b2 = β2
Locus is a2 x2 – y2 = b2 which is hyperbola.
If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for
exactly one value of a
no value of a
infinitely many values of a
exactly two values of a
B.
no value of a
S1 = x2 + y2 + 2ax + cy + a = 0 S2 = x2 + y2 – 3ax + dy – 1 = 0 Equation of radical axis of S1 and S2 S1 – S2 = 0 ⇒ 5ax + (c – d)y + a + 1 = 0 Given that 5x + by – a = 0 passes through P and Q
No real value to a
Sponsor Area
A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is
an ellipse
a circle
a hyperbola
a parabola
D.
a parabola
If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is
x2 + y2 – 3ax – 4by + (a2 + b2 – p2 ) = 0
2ax + 2by – (a2 – b2 + p2 ) = 0
x2 + y2 – 2ax – 3by + (a2 – b2 – p2 ) = 0
2ax + 2by – (a2 + b2 + p2 ) = 0
D.
2ax + 2by – (a2 + b2 + p2 ) = 0
Let the centre be (α, β)
∵It cut the circle x2 + y2 = p2 orthogonally
2(-α) × 0 + 2(-β) × 0 = c1 – p2
c1 = p2 Let equation of circle is x2 + y2 - 2αx - 2βy + p2 = 0
It pass through (a, b) ⇒ a2 + b2 - 2αa - 2βb + p2 = 0
Locus ∴ 2ax + 2by – (a2 + b2 + p2 ) = 0.
If the pair of lines ax2 + 2(a + b)xy + by2 = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then
3a2 – 10ab + 3b2 = 0
3a2 – 2ab + 3b2 = 0
3a2 + 10ab + 3b2 = 0
3a2 + 2ab + 3b2 = 0
D.
3a2 + 2ab + 3b2 = 0
If |z2-1|=|z|2+1, then z lies on
the real axis
an ellipse
a circle
the imaginary axis
B.
an ellipse
Given that
|z2- 1| = |z|2+ 2
|z2 + (-1)| = |z2| + |-1|
It shows that the origin, -1 and z2 lies on a line and z2 and -1 lies on one side of the origin, therefore
z2 is a negative number. Hence z will be purely imaginary. So we can say that z lies on y-axis.
If a circle passes through the point (a, b) and cuts the circle x2 +y2= 4 orthogonally, then the locus of its centre is
2ax +2by + (a2 +b2+4)=0
2ax +2by - (a2 +b2+4)=0
2ax -2by - (a2 +b2+4)=0
2ax -2by + (a2 +b2+4)=0
B.
2ax +2by - (a2 +b2+4)=0
Let the equation of circle is
x2 + y2 + 2gx + 2fy + c = 0
It cut the circle x2 + y2 = 4 orthogonally
if 2g.0 + 2f.0 = c - 4
⇒ c = 4
∴ Equation of circle is
x2 + y2 + 2gx + 2fy + 4 = 0
Q It passes through the point (a, b)
∴ a2 + b2 + 2ag + 2f + 4 = 0
Locus of centre (-g, -f) will be a2+ b2- 2xa - 2yb + 4 = 0
2ax + 2by - (a2 + b2 + 4) = 0
A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is
(x-p)2 = 4qy
(x-q)2 = 4py
(y-p)2 = 4qx
(y-p)2 = 4px
A.
(x-p)2 = 4qy
In a circle, AB is a diameter where the co-ordinate of A is (p, q) and let the co-ordinate of B is (x1 , y1 ).
Equation of circle in diameter form is (x - p)(x - x1 ) + (y - q)(y - y1 ) = 0
x2 - (p + x1 )x + px1 + y2 - (y1 + q)y + qy1 = 0
x2 - (p + x1 )x + y2 - (y1 + q)y + px1 + qy1 = 0
Since this circle touches X-axis
∴ y = 0
⇒ x2 - (p + x1 )x + px1 + qy1 = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x1 )2 = 4(px1 + qy1) p2 + x21 + 2px1
= 4px1 + 4qy1 x21 - 2px1 + p2 = 4qy1
Therefore the locus of point B is, (x - p)2 = 4qy
If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is
x2 + y2- 2x +2y -23 = 0
x2 - y2- 2x -2y -23 = 0
x2 - y2- 2x -2y +23 = 0
x2 + y2+ 2x +2y -23 = 0
A.
x2 + y2- 2x +2y -23 = 0
The lines 2x + 3y + 1 = 0 and 3x - y - 4 = 0 are diameters of circle.
On solving these equations we get x = 1, y = -1
Therefore the centre of circle = (1, -1) and circumference = 10 π
2πr = 10π
⇒ r = 5
∴ Equation of circle (x - x1 )2 + (y - y1 )2 = r2
(x - 1)2 + (y + 1)2 = 52
x2 + 1 - 2x + y2 + 2y + 1 = 25
x2 + y2 - 2x + 2y - 23 = 0
If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2+ 4ax = and x2+ 4ay = , then
d2 + (2b+3c)2 = 0
d2 +(3d+2c2) = 0
d2 + (2b-3c)2 = 0
d2 + (3b-2c)2 = 0
A.
d2 + (2b+3c)2 = 0
The equation of parabolas are y2 = 4ax and x2 = 4ay
On solving these we get x = 0 and x = 4a Also y = 0 and y = 4a
∴ The point of intersection of parabolas are A(0, 0) and B(4a, 4a). Also line 2bx + 3cy + 4d = 0 passes through A and B.
. ∴ d = 0 ............ (i)
or 2b . 4a + 3c . 4a + 4d = 0 2ab + 3ac + d = 0
a(2b + 3c) = 0 (Qd = 0)
⇒ 2b + 3c = 0 ............ (ii) On squaring equation (i) and (ii) and then adding, we get
d2 + (2b + 3c)2 = 0
The eccentricity of an ellipse, with its centre at the origin, is 1 /2 . If one of the directrices is x = 4, then the equation of the ellipse is
3x2 +4y2 = 1
3x2+ 4y2 = 12
4x2 +3y2 = 12
4x2+ 3y2 = 1
B.
3x2+ 4y2 = 12
Since the directrix is x = 4 then ellipse is parallel to X-axis.
⇒ a/e = 4
⇒ a = 4e = 4 x (1/2)
⇒ a = 2
Also we know that
b2 =a2 (1-e2)
b2 = 4(1-1/4) = 4 x 3/4
b2 = 3
therefore equation of ellipse is
Tangents are drawn to the hyperbola 4x2 – y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of △PTQ is
B.
Clearly, PQ is a chord of contact,
i.e., the equation of PQ is T = 0
=> y = –12
Solving with curve, 4x2 - y2 = 36
Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the
parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and,
then a value of tan θ is
4/3
1/2
2
3
C.
2
y2 = 16x
Tangent at P (16,6) is 2y = x + 16..... (1)
Normal at P (16,16) is y = -2x + 48... (2)
i.e., A is (-16,0); B is (24,0)
Now Centre of circle is (4,0)
Now, mPC = 4/3
mPB = -2
A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is
3x + 2y = 6xy
3x + 2y = 6
2x + 3y = xy
3x + 2y = xy
D.
3x + 2y = xy
Let the equation of line be (i)
(a) passes through the fixed point (2,3)
.... (ii)
P(a, 0), Q(0, b), O(0, 0), Let R(h, k),
Midpoint of OR is (h/2, k/2)
Midpoint of PQ is (a/2, b/2)
⇒h =a, k = b... (iii)
From (ii) & (iii)
Sponsor Area
Sponsor Area
