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Length of Chord

A1 and A2 are...

Question

$A_{1}$ and $A_{2}$ are the vertices of the conic $C_{1} : 4 (x - 3)^{2} + 9 (y - 2)^{2} - 36 = 0$ and point $P$ is moving in the plane such that $| P A_{1} - P A_{2} | = 3 \sqrt{2} .$ Then the locus of $P$ is another conic $C_{2} .$ If $D_{1}$ denotes the distance between foci of conic $C_{2},$ $D_{2}$ denotes the product of perpendicular distances from the points $A_{1}$ and $A_{2}$ upon any tangent drawn to the conic $C_{2}$ and $D_{3}$ denotes length of tangent drawn from any point on the auxiliary circle of conic $C_{1}$ to the auxiliary circle of the conic $C_{2},$ then ${(\frac{D_{1} D_{2}}{D_{3}^{2}})}^{2}$ is equal to

30

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20

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36

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48

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Solution

The correct option is C $36$
$C_{1} : \frac{(x - 3)^{2}}{9} + \frac{(y - 2)^{2}}{4} = 1$

$A_{1} = (6, 2)$ and $A_{2} = (0, 2)$
Given $| P A_{1} - P A_{2} | = 3 \sqrt{2}$
Clearly, locus of $P$ is a hyperbola for which
$A_{1} A_{2} = 2 a e = 6$
and $2 a = 3 \sqrt{2} \Rightarrow e = \sqrt{2}$
$∴$ Locus of $P$ is a rectangular hyperbola.
Equation of conic $C_{2}$ is $(x - 3)^{2} - (y - 2)^{2} = \frac{9}{2}$

Now, $D_{1} = 2 a e = 6$
We know that product of perpendiculars from foci upon any tangent is $(semi-conjugate axis)^{2} .$
$∴ D_{2} = b^{2} = \frac{9}{2}$
$D_{3} = \sqrt{9 - \frac{9}{2}} = \frac{3}{\sqrt{2}}$
$∴ {(\frac{D_{1} D_{2}}{D_{3}^{2}})}^{2} = 36$

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