Question

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. 126
• B. 135
• C. 145
• D. 116

Answer 1

Answer: (A)

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b) ; \frac{2 b^{2}}{a}=10$
$ \Rightarrow b^{2}=5 a \ldots(i)$
Now, $\phi(t)=\frac{5}{12}+t-t^{2}=\frac{8}{12}-\left(t-\frac{1}{2}\right)^{2}$
$\phi(t)_{\max }=\frac{8}{12}=\frac{2}{3}=e $
$\Rightarrow e^{2}=1-\frac{b^{2}}{a^{2}}=\frac{4}{9} \ldots$ (ii)
$\Rightarrow a^{2}=81 ($ from $(i) \&(i i))$
So, $a^{2}+b^{2}=81+45=126$