Question

Equation of the ellipse with eccentricity $ \frac{1}{2} $ and foci at $ (\pm 1, 0) $ is

• A. \( \frac{x^2}{3} + \frac{y^2}{4} = 1 \)
• B. \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \)
• C. \( \frac{x^2}{4} + \frac{y^2}{3} = \frac{4}{3} \)
• D. None of these

Hint:

For ellipses, use the relationship between eccentricity, semi-major, and semi-minor axes to find the equation. Remember the standard form is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Answer 1

The Correct Option is B

We are given an ellipse with eccentricity \( e = \frac{1}{2} \) and foci at \( (\pm 1, 0) \).
Step 1: Use the equation for an ellipse
The standard equation of an ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where \( a \) is the semi-major axis, and \( b \) is the semi-minor axis. The relationship between \( a \), \( b \), and the eccentricity \( e \) is: \[ e^2 = 1 - \frac{b^2}{a^2} \]
Step 2: Use the given eccentricity
Substituting \( e = \frac{1}{2} \), we get: \[ \left( \frac{1}{2} \right)^2 = 1 - \frac{b^2}{a^2} \] This simplifies to: \[ \frac{1}{4} = 1 - \frac{b^2}{a^2} \quad \Rightarrow \quad \frac{b^2}{a^2} = \frac{3}{4} \] Thus, \( b^2 = \frac{3}{4} a^2 \).
Step 3: Conclusion
Since the foci are at \( (\pm 1, 0) \), we can deduce that \( a = 2 \), so \( b = \sqrt{3} \). Thus, the correct equation of the ellipse is \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \), corresponding to option (b).