Suppose that the foci of the ellipse $\frac{x^2}{9}+ \frac{y^2}{5}=1$ are $\left(f_1, 0\right) and (f_2, 0)$ 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 $\left(f_1, 0\right)$ and $\left(2f_2, 0\right),$ respectively. Let $T_1$ be a tangent to $P_1$ which passes through $\left(2f_2, 0\right)$ 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 $\left(\frac{1}{m_1^2}+ m_2^2\right)$ is

Solution

$f_1 \equiv (2, 0)$
$f_2 \equiv (-2, 0)$
Parabolas are
$y^2= 8x$
$y^2= -16x$
$T_1 :y=mx+\frac{2}{m}$
$0= -4m+\frac{2}{m}$
$\Rightarrow m^2=\frac{1}{2}$
$T_2 :y=mx-\frac{4}{m}$
$2m-\frac{4}{m}=0$
$\Rightarrow m^2=2$
Now, $m_2^2+ \frac{1}{m_1^2}=2+2=4$

Hence, the correct answer is 4.