QJEEAdv15Ma6c1d: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma6c1d Solution: Suppose that the foci of the ellipse are and where and . Let…

Question

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Suppose that the foci of the ellipse

{{{x^2}} \over 9} + {{{y^2}} \over 5} = 1

are

\left( {{f_1},0} \right)

and

\left( {{f_2},0} \right)

where

{{f_1} &gt; 0}

and

{{f_2} &lt; 0}

. Let

{P_1}

and

{P_2}

be two parabolas with a common vertex at

(0,0)

and with foci at

\left( {{f_1},0} \right)

and

\left( 2{{f_2},0} \right)

, respectively. Let

{T_1}

be a tangent to

{P_1}

which passes through

\left( 2{{f_2},0} \right)

and

{T_2}

be a tangent to

{P_2}

which passes through

\left( {{f_1},0} \right)

. If

{m_1}

is the slope of

{T_1}

and

{m_2}

is the slope of

{T_2}

, then the value of

\left( {{1 \over {m_1^2}} + m_2^2} \right)

is

Accepted Answer

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

The foci are ( $\pm$ ae, 0) i.e. (2, 0) and ( $-$ 2, 0).

The parabola P₁ is ${y^2} = 8x$ and P₂ is ${y^2} = - 16x$

As tangent with slope m₁ to P₁ passes through ( $-$ 4, 0), we have

$y = {m_1}x + {2 \over {{m_1}}}$ giving $0 = - 4{m_1} + {2 \over {{m_1}}}$

i.e. $4m_1^2 = 2 \Rightarrow m_1^2 = {1 \over 2}$

Again for tangent with slope m₂ to P₂ passing through (2, 0), we have

$y = {m_2}x - {4 \over {{m_2}}} \Rightarrow 0 = 2{m_2} - {4 \over {{m_2}}}$

$\Rightarrow 2m_2^2 = 4$ $\therefore$ $m_2^2 = 2$

Thus, ${1 \over {m_1^2}} + m_2^2 = 2 + 2 = 4$

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