Question

$e$ is thc eccentricity of tho ellipse $4x^2+9y^2=36$ and $C$ is thc centre and $S$ is thc focus and $A$ is the vertex then $CS$:$SA=$

• A. $3-\sqrt{5}:\sqrt{5}$
• B. $\sqrt{5}:3-\sqrt{5}$
• C. $3+\sqrt{5}:\sqrt{5}$
• D. $\sqrt{5}:3+\sqrt{5}$

Answer 1

Answer: (B)

$\sqrt{5}:3-\sqrt{5}$

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$
$\frac{CS}{SA}=\frac{ae}{a-ae}=\frac{\sqrt{5}}{3-\sqrt{5}}$
Ans: B