Question

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. 29
• B. 31
• C. 32
• D. 34

Answer 1

Answer: (B)

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 a > b$
$e^{2}=1-\frac{b^{2}}{a^{2}}$
$\frac{1}{16}=1-\frac{b^{2}}{a^{2}}$
$\frac{b^{2}}{a^{2}}=1-\frac{1}{16}=\frac{15}{16}$
$\Rightarrow b^{2}=\frac{15}{16} a^{2}$
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\frac{16 \times \frac{2}{5}}{a^{2}}+\frac{9}{b^{2}}=1$
$\frac{32}{5 a^{2}}+\frac{9}{b^{2}}=1$
$\frac{32}{5 a^{2}}+\frac{9}{\frac{15}{16} a^{2}}=1$
$\frac{80}{5 a^{2}}=1$
$16= a ^{2}$
$b^{2}=15$