JEE PYQ: Ellipse Question 7
Question 7 - 2020 (04 Sep Shift 1)
Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, $\phi(t) = \frac{5}{12} + t - t^2$, then $a^2 + b^2$ is equal to:
(a) 145 (b) 116 (c) 126 (d) 135
Show Answer
Answer: (c) 126
Solution
$\frac{2b^2}{a} = 10 \Rightarrow b^2 = 5a$ …(i)
$\phi(t) = \frac{5}{12} + t - t^2$, $\phi’(t) = 1 - 2t = 0 \Rightarrow t = \frac{1}{2}$
$\phi(t)_{max} = \frac{5}{12} + \frac{1}{2} - \frac{1}{4} = \frac{2}{3}$
$e = \frac{2}{3}$
$b^2 = a^2(1 - e^2) \Rightarrow 5a = \frac{5a^2}{9} \Rightarrow a = 9$
$a^2 = 81$, $b^2 = 45$
$a^2 + b^2 = 126$
BETA
