Let the eccentricity of an ellipse frac{x^{2} | vedclass

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(A) Given the ellipse equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a>b$ and eccentricity $e = \frac{1}{4}$.We know $e^{2} = 1 - \frac{b^

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$31$

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(A) Given the ellipse equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a>b$ and eccentricity $e = \frac{1}{4}$.We know $e^{2} = 1 - \frac{b^{2}}{a^{2}}$,so $\frac{1}{16} = 1 - \frac{b^{2}}{a^{2}}$,which gives $\frac{b^{2}}{a^{2}} = \frac{15}{16}$ or $b^{2} = \frac{15}{16}a^{2}$.The ellipse passes through $\left(-4 \sqrt{\frac{2}{5}}, 3ight)$,so $\frac{(-4 \sqrt{2/5})^{2}}{a^{2}} + \frac{3^{2}}{b^{2}} = 1$.Substituting the values: $\frac{16 	imes (2/5)}{a^{2}} + \frac{9}{b^{2}} = 1 \Rightarrow \frac{32}{5a^{2}} + \frac{9}{b^{2}} = 1$.Substitute $b^{2} = \frac{15}{16}a^{2}$: $\frac{32}{5a^{2}} + \frac{9 	imes 16}{15a^{2}} = 1$.$\frac{32}{5a^{2}} + \frac{144}{15a^{2}} = 1$ $\Rightarrow \frac{96 + 144}{15a^{2}} = 1$ $\Rightarrow \frac{240}{15a^{2}} = 1$.$16 = a^{2}$.Then $b^{2} = \frac{15}{16} 	imes 16 = 15$.Thus,$a^{2} + b^{2} = 16 + 15 = 31$.

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

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