Question

Let the eccentricity of an ellipse $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$, $a>b$, be $\frac 14$. If this ellipse passes through the point $(−4\sqrt {\frac 25},3)$, then $a^2 + b^2$ is equal to :

• A. 29
• B. 31
• C. 32
• D. 34

Answer 1

Answer: (B)

31

Answer 2

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

$⇒ \frac {(−4\sqrt {\frac 25})^2}{a^2}+\frac {3^2}{b^2}=1$

$⇒ \frac {32}{a^2}+\frac {9}{b^2}=1$ ....(i)

$a^2(1-e^2)=b^2$
$⇒ a^2(1-\frac {1}{16}) = b^2$
⇒ 15a^2 = 16 b^2$$ ⇒ a^2 = \frac {16b^2}{15}
From (i)
$\frac {6}{b^2}+\frac {9}{b^2}=1$
$⇒ b^2 = 15$
and $a^2 = 16$
$⇒ a^2+b^2 = 15+16$ $= 31$

So, the correct option is (B): $31$