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)

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