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Conic Section

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Question
CBSEENMA11015637
Wired Faculty App

If space straight omega space equals space fraction numerator straight z over denominator straight z minus begin display style 1 third end style straight i space end fraction space and space vertical line straight omega vertical line space equals 1 comma space then space straight z space lies space on

  • an ellipse

  • a circle

  • a straight line

  • a parabola

Solution

C.

a straight line

As given WiredFaculty ⇒ 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.

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Question
CBSEENMA11015647
Wired Faculty App

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

Solution

D.

a parabola

WiredFaculty
Equation of circle with centre (0, 3) and radius 2 is
x2 + (y – 3)2 = 4.
Let locus of the variable circle is (α, β)
∵It touches x-axis. ∴ It equation (x - α) 2 + (y - β) 2 = β2 Circles touch externally
therefore space square root of straight alpha squared plus left parenthesis straight beta minus 3 right parenthesis squared end root space equals 2 plus straight beta
α2 + (β - 3)2 = β2 + 4 + 4β α2 = 10(β - 1/2)
∴ Locus is x2 = 10(y – 1/2) which is parabola.

Question
CBSEENMA11015567
Wired Faculty App

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

Solution

A.

8/3

Major axis is along x-axis.
WiredFaculty

Question
CBSEENMA11015568
Wired Faculty App

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at 

  • (0, 2)

  • (1, 0)

  • (0,1)

  • (2,0)

Solution

B.

(1, 0)

vertex (0,1) 
WiredFaculty

Question
CBSEENMA11015683
Wired Faculty App

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

Solution

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

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