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Suppose that the foci of the ellipse $\mathrm{\frac{x^2}{9}+\frac{y^2}{5}=1}$ are $\mathrm{\left(f_1, 0\right) and \left(f_2, 0\right)}$ where $\mathrm{f_1>0 \ and \ f_2<0}$ . Let $\mathrm{P_1 \ and \ P_2}$ be two parabolas with a common vertex at (0,0) and with foci at $\mathrm{\left(f_1, 0\right) \text { and }\left(2 f_2, 0\right)}$ respectively. Let $\mathrm{T_1}$ be a tangent to $\mathrm{P_1}$ which passes through $\mathrm{\left(2 f_2, 0\right)}$ and $\mathrm{T_2}$ be a tangent to $\mathrm{P_2}$ which passes through $\mathrm{ \left(f_1, 0\right)}$ . If $\mathrm{m_1}$ is the slope of $\mathrm{T_1}$ and $\mathrm{m_2}$ is the slope of $\mathrm{T_2}$ , then the value of $\mathrm{ \left(\frac{1}{m_1^2}+m_2^2\right)}$ is
Option: 1
4

Option: 2
3

Option: 3
2

Option: 4
1

Answers (1)

$\mathrm{e^2=1-\frac{b^2}{a^2}=1-\frac{5}{9}=\frac{4}{9}}$

The foci are (±ae, 0) i.e. (2, 0) and (–2, 0).

$\mathrm{\text { The parabola } P_1 \text { is } y^2=8 x \text { and } P_2 \text { is } y^2=-16 x}$

As tangent with slope $\mathrm{m_1 \text { to } P_1}$ passes through (–4, 0),

$\mathrm{\text { we have } y=m_1 x+\frac{2}{m_1} \text { giving } 0=-4 m_1+\frac{2}{m_1}}$

$\mathrm{\text { i.e. } 4 m_1^2=2 \Rightarrow m_1^2=\frac{1}{2}}$

Again for tangent with slope $\mathrm{m_2 \text { to } P_2}$ passing through (2, 0),

$\mathrm{\text { we have } y=m_2 x-\frac{4}{m_2} \Rightarrow 0=2 m_2-\frac{4}{m_2}}$

$\mathrm{\Rightarrow \quad 2 m_2^2=4 \quad \therefore m_2^2=2}$

$\mathrm{\text { Thus, } \frac{1}{m_1^2}+m_2^2=2+2=4}$

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