Conics related problem

Question

Let $A_1$ and $A_2$ are the vertices of the conic $C_1 : 4(x – 3)^2 + 9(y – 2)^2 – 36 = 0$ and a point P is moving in the plane such that $|PA_1-PA_2|=3\sqrt{2}$ , then locus of P is another conic $C_2$. If $D_1$ denotes distance between foci of conic $C_2$. $D_2$ denotes product of the perpendiculars from the points $A_1$ , $A_2$ upon any tangent drawn to conic $C_2$ and $D_3$ denotes length of the tangent drawn from any point on auxiliary circle of conic $C_1$ to the auxiliary circle of the conic $C_2$, then $\left(\frac{D_1.D_2}{D_3^2}\right)^2$ is equal to____________

My approach is as follow ${C_1}:\frac{{{{\left( {x - 3} \right)}^2}}}{9} + \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1$

The locus of $C_2$ represent a hyperbola whose focus are $A_{1}$ and $A_2$

$A_1.A_2=a_{hyperbola}e_{hyperbola}=D_1$, where $2a_{hyperbola}=3\sqrt{2}$

I am not able to approach further

Answer 1

First shift both $C_1$ and $C_2$ so that they're centered at the origin. This does not affect $D_1, D_2, D_3$.

$D_1$ is straight forward to compute. From the equation of $C_1$ it follows that $D_1 = 6$.

Next, you have to find $a$ and $b$ of $C_2$ whose equation is

$$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$

The equation to use is $2 a = 3\sqrt{2}$, hence, $a = \dfrac{3}{\sqrt{2}}$

Then to find $b$, use $ c = 3 = a e = a \sqrt{1 + \frac{b^2}{a^2} } $.

Hence, $b = a = \dfrac{3}{\sqrt{2}} $

So the equation of the shifted $C_2$ is

$$ x^2 - y^2 = \dfrac{9}{2} $$

Next, point a point $(x_0, y_0)$ on the hyperbola and find the tangent line. From implicit differentiation,

$y' = \dfrac{x_0}{y_0}$

Hence the equation of the tangent is $ y_0 (y - y_0) - x_0( x - x_0) = 0 $

Simplifying, this becomes $ y_0 y - x_0 x + \frac{9}{2} = 0 $

Now find the distances of the shifted vertices $(-3, 0)$ and $(3, 0)$ from this line and multiply them, we have

$p_1 = \dfrac{ 3 x_0 + \dfrac{9}{2} }{ \sqrt{x_0^2 + y_0^2} } $

$p_2 = - \dfrac{ -3 x_0 + \dfrac{9}{2}}{\sqrt{x_0^2 + y_0^2}}$

So that, the product $p_1 p_2 $ is

$D_2 = p_1 p_2 = \dfrac{ 9 x_0^2 - \dfrac{81}{4} } { x_0^2 + y_0^2 } $

But $y_0^2 = x_0^2 - \dfrac{9}{2}$, so

$D_2 = \dfrac{ \dfrac{9}{2} ( 2 x_0^2 - \dfrac{9}{2} ) }{ 2 x_0^2 - \dfrac{9}{2}} = \dfrac{9}{2} $

Finally, $D_3$ is the length of the chord in the auxiliary circle of $C_1$ touching the auxiliary circle of $C_2$. So it length is

$D_3 = 2 \sqrt{ 3^2 - a^2 } = 2 \sqrt{ 9 - \dfrac{9}{2} } = \dfrac{6}{\sqrt{2}}$

Hence,

$\left( \dfrac{D_1 D_2}{D_3^2} \right)^2 = \left( \dfrac{6 \times \dfrac{9}{2}}{18} \right)^2 = \left( \dfrac{ 27 }{18} \right)^2 = \left( \dfrac{3}{2} \right)^2 = \dfrac{9}{4} $.