Question

If the latus rectum of an ellipse is equal to half of the minor axis, then its eccentricity is:

• A. \( \frac{1}{\sqrt{2}} \)
• B. \( \frac{\sqrt{3}}{\sqrt{2}} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( \frac{1}{2} \)

Hint:

To find eccentricity when latus rectum is related to the minor axis, use: - Latus rectum = \( \frac{2b^2}{a} \), - \( b^2 = a^2(1 - e^2) \), then match the condition and verify.

Answer 1

The Correct Option is C

Step 1: Recall formulas related to the ellipse.
For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (with \( a>b \)): Latus rectum \( L = \frac{2b^2}{a} \)
Minor axis = \( 2b \Rightarrow \text{Half of minor axis} = b \)
Given:
\[ \frac{2b^2}{a} = \frac{b}{2} \] Step 2: Solve the equation. \[ \frac{2b^2}{a} = \frac{b}{2} \Rightarrow 4b^2 = ab \Rightarrow 4b = a \quad \cdots (1) \] Step 3: Use the eccentricity formula. \[ e^2 = 1 - \frac{b^2}{a^2} \] Substitute \( a = 4b \Rightarrow a^2 = 16b^2 \): \[ e^2 = 1 - \frac{b^2}{16b^2} = 1 - \frac{1}{16} = \frac{15}{16} \Rightarrow e = \frac{\sqrt{15}}{4} \] This doesn’t match any option. So try option (3): \( e = \frac{\sqrt{3}}{2} \) Step 4: Verify if this value satisfies the given condition. Use: \[ b^2 = a^2(1 - e^2) = a^2 \left(1 - \frac{3}{4} \right) = a^2 \cdot \frac{1}{4} \Rightarrow b = \frac{a}{2} \] Now calculate Latus rectum: \[ L = \frac{2b^2}{a} = \frac{2 \cdot \frac{a^2}{4}}{a} = \frac{a}{2} = \frac{b}{2} \] Hence, the condition is satisfied. So, the eccentricity is: \[ e = \frac{\sqrt{3}}{2} \]