Let the eccentricity of an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,a > b,$ be $\frac{1}{4}.$ If this ellipse passes through the point $\left( { - 4\sqrt {\frac{2}{5}} ,3} \right),$ then ${a^2} + {b^2}$ is equal to:

JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

Let the eccentricity of an ellipse

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

\frac{1}{4}.

If this ellipse passes through the point

\left( { - 4\sqrt {\frac{2}{5}} ,3} \right),

then

{a^2} + {b^2}

is equal to:

View answer and explanation

Correct option: B

Option B (31) is correct because by substituting the given point and eccentricity into the ellipse equation, we can solve for a² + b².

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r

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Q

P

O