Suppose that the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ are $(f_{1}, 0)$ and $(f_{2}, 0)$ where $f_{1} > 0$ and $f_{2} < 0.$
Let $P_{1}$ and $P_{2}$ be two parabolas with a common vertex at $(0, 0)$ and with foci at $(f_{1}, 0)$ and $(2 f_{2}, 0),$ respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2 f_{2}, 0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1}, 0) .$ If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$ then the value of $(\frac{1}{m_{1}^{2}} + m_{2}^{2})$ is

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Solution

$W e h a v e e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{5}{9} = \frac{4}{9} \Rightarrow e = \frac{2}{3}$
$∴ (f_{1}, 0) = (2, 0) and (f_{2}, 0) = (- 2, 0)$
$\Rightarrow P_{1} \Rightarrow y^{2} = 8 x a n d P_{2} \Rightarrow y^{2} = - 16 x$
$∴ E q u a t i o n s o f \tan g e n t s T_{1} a n d T_{2} s a t i s f y t h e f o l l o w i n g c o n d i t i o i n s$
$0 = - 4 m_{1} + \frac{2}{m_{1}}, 2 m_{2} - \frac{4}{m_{2}} = 0$
$\Rightarrow m_{1}^{2} = \frac{1}{2}, m_{2}^{2} = 2$
$\Rightarrow \frac{1}{m_{1}^{2}} = 2, \frac{1}{m_{2}^{2}} = 4$ '
$\Rightarrow \frac{1}{m_{1}^{2}} + m_{2}^{2} = 4$

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