Slope Ratio Sum of Tangents to Parabolas from Ellipse Foci

To find the value of $\left(\frac{1}{m_1^2} + m_2^2\right)$, we begin by determining the foci of the given ellipse.

Step 1: Find the foci of the ellipse

The equation of the ellipse is given by: $\frac{x^2}{9} + \frac{y^2}{5} = 1$

Here, $a^2 = 9$ and $b^2 = 5$. The eccentricity $e$ of the ellipse is calculated as: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

The foci of the ellipse are at $(\pm ae, 0) = \left(\pm 3 \cdot \frac{2}{3}, 0\right) = (\pm 2, 0)$.

Since $f_1 > 0$ and $f_2 < 0$, we have: $f_1 = 2 \quad \text{and} \quad f_2 = -2$

Thus, the points are $(f_1, 0) = (2, 0)$ and $(2f_2, 0) = (-4, 0)$.

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Step 2: Determine equation and slope of tangent $T_1$ for Parabola $P_1$

The parabola $P_1$ has a vertex at $(0,0)$ and focus at $(f_1, 0) = (2, 0)$. The standard equation of $P_1$ is: $y^2 = 4(2)x \implies y^2 = 8x$

The slope form of a tangent to $y^2 = 4ax$ is $y = mx + \frac{a}{m}$. Here $a = 2$, so the tangent line $T_1$ with slope $m_1$ is: $y = m_1 x + \frac{2}{m_1}$

Since $T_1$ passes through $(2f_2, 0) = (-4, 0)$: $0 = m_1(-4) + \frac{2}{m_1}$ $4m_1 = \frac{2}{m_1} \implies m_1^2 = \frac{1}{2}$

Therefore: $\frac{1}{m_1^2} = 2$

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Step 3: Determine equation and slope of tangent $T_2$ for Parabola $P_2$

The parabola $P_2$ has a vertex at $(0,0)$ and focus at $(2f_2, 0) = (-4, 0)$. The standard equation of $P_2$ is: $y^2 = -4(4)x \implies y^2 = -16x$

The tangent line $T_2$ with slope $m_2$ to $y^2 = -4ax$ (where $a = 4$) is: $y = m_2 x - \frac{4}{m_2}$

Since $T_2$ passes through $(f_1, 0) = (2, 0)$: $0 = m_2(2) - \frac{4}{m_2}$ $2m_2 = \frac{4}{m_2} \implies m_2^2 = 2$

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Step 4: Calculate the required expression

Substituting the values of $\frac{1}{m_1^2}$ and $m_2^2$ into the target expression: $\frac{1}{m_1^2} + m_2^2 = 2 + 2 = 4$