Suppose that the foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ are $\left(f_{1}, 0\right)$

Question

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

Professor PC Thomas Classes · Answer

(4) : $e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{5}{9}=\frac{4}{9}$ The foci are $( \pm a e, 0)$ i.e. $(2,0)$ and $(-2,0)$.\ The parabola $P_{1}$ is $y^{2}=8 x$ and $P_{2}$ is $y^{2}=-16 x$ As tangent with slope $m_{1}$ to $P_{1}$ passes through $(-4,0)$, we have\ $y=m_{1} x+\frac{2}{m_{1}}$ giving $0=-4 m_{1}+\frac{2}{m_{1}}$\ i.e. $4 m_{1}^{2}=2 \Rightarrow m_{1}^{2}=\frac{1}{2}$ Again, for tangent with slope $m_{2}$ to $P_{2}$ passing through $(2,0)$, we have\ $y=m_{2} x-\frac{4}{m_{2}} \Rightarrow 0=2 m_{2}-\frac{4}{m_{2}}$\ $\Rightarrow \quad 2 m_{2}{ }^{2}=4 \quad 	herefore m_{2}{ }^{2}=2$\ Thus, $\frac{1}{m_{1}^{2}}+m_{2}^{2}=2+2=4$\