JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 8)
Let the eccentricity of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be reciprocal to that of the ellipse $${x^2} + 4{y^2} = 4$$. If the hyperbola passes through a focus of the ellipse, then
the equation of the hyperbola is $${{{x^2}} \over 3} - {{{y^2}} \over 2} = 1$$
a focus of the hyperbola is $$(2, 0)$$
theeccentricity of the hyperbola is $$\sqrt {{5 \over 3}} $$
The equation of the hyperbola is $${x^2} - 3{y^2} = 3$$
Explanation
Ellipse is $${{{x^2}} \over {{2^2}}} + {{{y^2}} \over {{1^2}}} = 1$$.
$${1^2} = {2^2}(1 - {e^2}) \Rightarrow e = {{\sqrt 3 } \over 2}$$
Therefore, the eccentricity of the hyperbola is
$${2 \over {\sqrt 3 }} \Rightarrow {b^2} = {a^2}\left( {{4 \over 3} - 1} \right) \Rightarrow 3{b^2} = {a^2}$$
Foci of the ellipse are $$(\sqrt 3 ,\,0)$$ and $$( - \sqrt 3 ,\,0)$$.
Hyperbola passes through $$( \sqrt 3 ,\,0)$$.
$${3 \over {{a^2}}} = 1 \Rightarrow {a^2} = 3$$ and $${b^2} = 1$$
Therefore, the equation of hyperbola is $${x^2} - 3{y^2} = 3$$.
Focus of hyperbola is $$(ae,\,0) \equiv \left( {\sqrt 3 \times {2 \over {\sqrt 3 }},\,0} \right) \equiv (2,0)$$.
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