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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}

Posted by

Sumit Saini

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