Let eccentricity of an ellipse frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1, a>b , be frac{1}{4} . If

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10-2. Parabola, Ellipse, Hyperbola

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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

A

$31$

B

$29$

C

$32$

D

$34$

(JEE MAIN-2022)

Solution

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

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

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

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

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

$\frac{16 \times \frac{2}{5}}{a^{2}}+\frac{9}{b^{2}}=1$

$\frac{32}{5 a^{2}}+\frac{9}{b^{2}}=1$

$\frac{32}{5 a^{2}}+\frac{9}{\frac{15}{16} a^{2}}=1$

$\frac{80}{b^{2}}=1$

Standard 11

Mathematics

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