Suppose that the foci of the ellipse frac{x^2 | vedclass

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(D) For the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,$a^2 = 9$ and $b^2 = 5$. The eccentricity $e = \sqrt{1 - \frac{5}{9}} = \frac{2}{3}$.Foci are

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$4$

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(D) For the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,$a^2 = 9$ and $b^2 = 5$. The eccentricity $e = \sqrt{1 - \frac{5}{9}} = \frac{2}{3}$.Foci are $(\pm ae, 0) = (\pm 3 	imes \frac{2}{3}, 0) = (\pm 2, 0)$. Thus $f_1 = 2$ and $f_2 = -2$.Parabola $P_1$ has vertex $(0,0)$ and focus $(f_1, 0) = (2, 0)$,so its equation is $y^2 = 4(2)x = 8x$.Parabola $P_2$ has vertex $(0,0)$ and focus $(2f_2, 0) = (-4, 0)$,so its equation is $y^2 = 4(-4)x = -16x$.Tangent $T_1$ to $y^2 = 8x$ passes through $(2f_2, 0) = (-4, 0)$. The equation of a tangent to $y^2 = 4ax$ with slope $m_1$ is $y = m_1x + \frac{a}{m_1}$. Here $a=2$,so $y = m_1x + \frac{2}{m_1}$.Substituting $(-4, 0)$: $0 = -4m_1 + \frac{2}{m_1}$ $\Rightarrow 4m_1 = \frac{2}{m_1}$ $\Rightarrow m_1^2 = \frac{1}{2}$ $\Rightarrow \frac{1}{m_1^2} = 2$.Tangent $T_2$ to $y^2 = -16x$ passes through $(f_1, 0) = (2, 0)$. The equation of a tangent to $y^2 = -4ax$ with slope $m_2$ is $y = m_2x - \frac{a}{m_2}$. Here $a=4$,so $y = m_2x - \frac{4}{m_2}$.Substituting $(2, 0)$: $0 = 2m_2 - \frac{4}{m_2}$ $\Rightarrow 2m_2 = \frac{4}{m_2}$ $\Rightarrow m_2^2 = 2$.Therefore,$\frac{1}{m_1^2} + m_2^2 = 2 + 2 = 4$.

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