1. Find the foci of the ellipse

The ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$ So, $a^2=9,\quad b^2=5$ Hence, $c^2=a^2-b^2=9-5=4 \implies c=2$ Therefore the foci are at $(\pm 2,0)$.

Given: $f_1>0,\; f_2<0$ So, $f_1=2,\qquad f_2=-2$

2. Write equations of the two parabolas

Parabola $P_1$

It has vertex at $(0,0)$ and focus at $(f_1,0)=(2,0)$. A parabola with vertex at origin and focus $(a,0)$ has equation $y^2=4ax$ Here $a=2$, so $P_1: y^2=8x$

Parabola $P_2$

It has vertex at $(0,0)$ and focus at $(2f_2,0)=(2\cdot -2,0)=(-4,0)$. A parabola with vertex at origin and focus $(-a,0)$ has equation $y^2=-4ax$ Here $a=4$, so $P_2: y^2=-16x$

3. Find slope $m_1$ of tangent $T_1$ to $P_1$ passing through $(2f_2,0)=(-4,0)$

For parabola $y^2=4ax$ A tangent with slope $m$ is $y=mx+\frac{a}{m}$

For $P_1: y^2=8x$, we have $4a=8\Rightarrow a=2$. So tangent is $y=mx+\frac{2}{m}$

Since it passes through $(-4,0)$, $0=m(-4)+\frac{2}{m}$ Multiply by $m$: $-4m^2+2=0$ $4m^2=2$ $m^2=\frac12$ Thus, $\frac{1}{m_1^2}=2$

4. Find slope $m_2$ of tangent $T_2$ to $P_2$ passing through $(f_1,0)=(2,0)$

For parabola $y^2=-4ax$ A tangent with slope $m$ is $y=mx-\frac{a}{m}$

For $P_2: y^2=-16x$, we have $4a=16\Rightarrow a=4$. So tangent is $y=mx-\frac{4}{m}$

Since it passes through $(2,0)$, $0=2m-\frac{4}{m}$ Multiply by $m$: $2m^2-4=0$ $m^2=2$ Thus, $m_2^2=2$

5. Compute the required value

We need $\frac{1}{m_1^2}+m_2^2$ Substitute: $\frac{1}{m_1^2}=2,\qquad m_2^2=2$ Therefore, $\frac{1}{m_1^2}+m_2^2=2+2=4$

6. Comparison with stored answer

Derived answer = $4$. Stored correct answer = $4$. They agree.