JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

Let

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

be a given ellipse, length of whose latusrectum 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

View answer and explanation

Correct option: B

126

Option B (126) is correct because the values of a and b are calculated using the latus rectum and eccentricity, and then

a^2 + b^2

is computed.

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