JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 11)
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 $${{f_1} > 0}$$ and $${{f_2} < 0}$$. 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
Answer
4
Explanation
$${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {5 \over 9} = {4 \over 9}$$
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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