For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, if the $y$-axis is the minor axis and the length of the latus rectum is one half of the minor axis, then the eccentricity is
- A. $\frac{1}{\sqrt{2}}$
- B. $\frac{1}{2}$
- C. $\frac{\sqrt{3}}{2}$
- D. $\frac{3}{4}$
✅ Correct Answer: $\frac{\sqrt{3}}{2}$
Explanation
Latus rectum $= \frac{2b^2}{a} = \frac{1}{2}(2b) \Rightarrow b^2 = \frac{a^2}{4}$. Thus $e = \sqrt{1 - \frac{b^2}{a^2}} = \frac{\sqrt{3}}{2}$.
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