Let the eccentricity of the hyperbola $H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ be $\sqrt {{5 \over 2}}$ and length of its latus rectum be $6\sqrt 2$. If $y = 2x + c$ is a tangent to the hyperbola H, then the value of c2 is equal to :
Solution
<p>$$1 + {{{b^2}} \over {{a^2}}} = {5 \over 2} \Rightarrow {{{b^2}} \over {{a^2}}} = {3 \over 2}$$</p>
<p>${{2{b^2}} \over a} = 6\sqrt 2 \Rightarrow 2.\,{3 \over 2}.\,a = 6\sqrt 2$</p>
<p>$\Rightarrow a = 2\sqrt 2 ,\,{b^2} = 12$</p>
<p>${c^2} = {a^2}{m^2} - {b^2} = 8.4 - 12 = 20$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
This question is part of PrepWiser's free JEE Main question bank. 146 more solved questions on Conic Sections are available — start with the harder ones if your accuracy is >70%.