JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q13

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MathematicsSingle correct+4 / -1Question 13 of 90

Let the eccentricity of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , $a>b$ , be $\dfrac14$ . If this ellipse passes through the point $\left(-4\sqrt{\dfrac25},3\right)$ , then $a^2+b^2$ is equal to :

Answer and worked solution

Official answer: B

Eccentricity. $b^2=a^2\left(1-\frac1{16}\right)=\frac{15}{16}a^2$ .

Through the point. $\dfrac{32/5}{a^2}+\dfrac9{b^2}=\dfrac{32}{5a^2}+\dfrac{48}{5a^2}=\dfrac{16}{a^2}=1$ , so $a^2=16$ , $b^2=15$ .

Sum 31.

(B)

Topics:Ellipse

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JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 13