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Ellipse

Section: Mathematics

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Question : 9 of 117

Marks: +1, -0

Let the length of the latus rectum of an ellipse

\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1,(a>b)

, be 30 . If its eccentricity is the maximum value of the function

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

(a^2+b^2)

[24 Jan 2026 Shift 2]

276
516
256
496

Solution:

Concept:

Latus rectum of ellipse

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

with

a>b

is

\frac{2b^2}{a}

.

Eccentricity

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

.

Maximum of quadratic

f(t)=At^2+Bt+C

occurs at

t=-\frac{B}{2A}

when

A<0

.

Explanation:

Given latus rectum length = 30.

So

\frac{2b^2}{a}=30

implies

b^2=15a

.

Function

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

.

Coefficient of

t^2

is negative, so maximum at

t=-\frac{2}{2(-1)}=1

.

Maximum value

f(1)= -1+2-\frac{3}{4}= \frac{1}{4}

.

Thus eccentricity

e=\frac{1}{4}

.

For ellipse,

e^2 = 1-\frac{b^2}{a^2} = \frac{1}{16}

.

Substitute

b^2=15a

:

1-\frac{15a}{a^2}=\frac{1}{16}

.

So

\frac{15}{a}=1-\frac{1}{16}=\frac{15}{16}

.

Hence

a=16

.

Then

b^2=15\times 16 = 240

and

a^2=256

.

Thus

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

.

Answer:

Option D: 496.

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