Let the hyperbola $H:{{{x^2}} \over {{a^2}}} - {y^2} = 1$ and the ellipse $E:3{x^2} + 4{y^2} = 12$ be such that the length of latus rectum of H is equal to the length of latus rectum of E. If ${e_H}$ and ${e_E}$ are the eccentricities of H and E respectively, then the value of $12\left( {e_H^2 + e_E^2} \right)$ is equal to ___________.
Answer (integer)
42
Solution
<p>$\because$ $H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 1} = 1$</p>
<p>$\therefore$ Length of latus rectum $= {2 \over a}$</p>
<p>$E:{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$</p>
<p>Length of latus rectum $= {6 \over 2} = 3$</p>
<p>$\because$ ${2 \over a} = 3 \Rightarrow a = {2 \over 3}$</p>
<p>$\therefore$ $$12\left( {e_H^2 + e_E^2} \right) = 12\left( {1 + {9 \over 4}} \right) + \left( {1 - {3 \over 4}} \right) = 42$$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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