(A) 276

(B) 516

(C) 256

(D) 496

Correct Answer: 496

Explanation

For the ellipse

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$

the length of the latus rectum is given by

$$ \text{Latus rectum}=\frac{2b^2}{a} $$

Given:

$$ \frac{2b^2}{a}=30 \Rightarrow b^2=15a $$

The eccentricity of the ellipse is

$$ e=\sqrt{1-\frac{b^2}{a^2}} $$

Given that eccentricity equals the maximum value of

$$ f(t)=-\frac{3}{4}+2t-t^2 $$

This quadratic function has maximum value at

$$ t=\frac{2}{2}=1 $$

Maximum value:

$$ f_{\max}=-\frac{3}{4}+2(1)-1^2=\frac{1}{4} $$

Hence,

$$ e=\frac{1}{4} $$

So,

$$ \sqrt{1-\frac{b^2}{a^2}}=\frac{1}{4} $$ $$ 1-\frac{b^2}{a^2}=\frac{1}{16} $$ $$ \frac{b^2}{a^2}=\frac{15}{16} $$

Using \( b^2=15a \):

$$ \frac{15a}{a^2}=\frac{15}{16} $$ $$ a=16 $$

Then,

$$ b^2=15a=240 $$

Finally,

$$ a^2+b^2=16^2+240=256+240=496 $$

Hence, the correct answer is 496.

Related JEE Main Questions