Question

If $B$ is the end point of minor axis of the ellipse $b^2x^2+a^2y^2=a^2b^2\ (a>b)$ and $S$ and $S'$ are the foci of the ellipse such that $\triangle BSS'$ is an equilateral triangle, then the eccentricity $e$ is

• A. $\dfrac{1}{2}$
• B. $\dfrac{1}{3}$
• C. $\dfrac{3}{5}$
• D. $\dfrac{4}{5}$

Hint:

For ellipse problems involving geometry, always use the distance formula along with standard ellipse parameters.

Answer 1

The Correct Option is A

Step 1: Identifying key elements of the ellipse.
The given ellipse is \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a>b \] The foci are at $(\pm c,0)$ where $c^2=a^2-b^2$. The end point of the minor axis is $B(0,b)$. 
Step 2: Using the equilateral triangle condition. 
Since $\triangle BSS'$ is equilateral, \[ BS = SS' \] But $SS' = 2c$. 
Step 3: Finding $BS$. 
\[ BS = \sqrt{c^2+b^2} \] Equating: \[ \sqrt{c^2+b^2} = 2c \] 
Step 4: Solving for $c$ in terms of $b$. 
\[ c^2+b^2=4c^2 \Rightarrow 3c^2=b^2 \Rightarrow c^2=\frac{b^2}{3} \] 
Step 5: Finding eccentricity. 
\[ e=\frac{c}{a},\quad a^2=b^2+c^2=\frac{4b^2}{3} \] \[ e=\frac{\frac{b}{\sqrt{3}}}{\frac{2b}{\sqrt{3}}}=\frac{1}{2} \] 
Step 6: Conclusion. 
The eccentricity of the ellipse is $\dfrac{1}{2}$.