Question

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ be given ellipse, length of whose latus rectum is $10.$ If its eccentricity is the maximum value of the function, $\varphi \left(t\right)=\frac{5}{12}+t-t^2,\text{then}{a}^2+b^2$ is equal to

• A. $135$
• B. $116$
• C. $126$
• D. $145$

Answer 1

The correct option is C $126$

Given: $\varphi \left(t\right)=\frac{5}{12}-{(t-\frac{1}{2})}^2+\frac{1}{4}$
$\Rightarrow \varphi \left(t\right)=\frac{8}{12}-{(t-\frac{1}{2})}^2$
$\therefore \varphi (t{)}_{max}=\frac{2}{3}=e$
We know that
$e^2=1-\frac{b^2}{a^2}\Rightarrow \frac{b^2}{a^2}=\frac{5}{9}\cdots \left(1\right)$
Also, $L.R.=\frac{2b^2}{a}=10\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$a=9$ and $b^2=45$
$\therefore a^2+b^2=81+45=126$