A parabola has vertex $(2, 3)$, equation of directrix is $2x – y = 1$ and equation of ellipse is $\frac {x^2}{a^2} +\frac {y^2}{b^2} =1$, $e=\frac {1}{\sqrt 2}$ and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is

Question

Let the length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\dfrac{3}{4}+2t-t^2$, then $(a^2+b^2)$ is equal to

Answer 1

Option (D) is correct: $ \frac {4352}{25} $.

A parabola has vertex \((2, 3)\), equation of directrix is \(2x – y = 1\) and equation of ellipse is \(\frac {x^2}{a^2} +\frac {y^2}{b^2} =1\), \(e=\frac {1}{\sqrt 2}\) and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is

The Correct Option is D

Solution and Explanation

Top Questions on Ellipse

Questions Asked in JEE Main exam