Question

The equation of the ellipse with foci at ($\pm$3, 0) and the eccentricity as 1/3 is:

• A. $\frac{x^{2}}{81} + \frac{y^{2}}{72} = 1$
• B. $\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$
• C. $\frac{x^{2}}{8} + \frac{y^{2}}{9} = 1$
• D. $\frac{x^{2}}{3} + \frac{y^{2}}{2} = 1$

Hint:

If foci are $(\pm c, 0)$, then $a$ must be greater than $c$. Here $a=9 > c=3$, confirming our steps.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The foci of an ellipse on the x-axis are given by \( (\pm ae, 0) \). By using the given foci and eccentricity, we can solve for the semi-major axis \( a \) and then find the semi-minor axis \( b \).
Step 2: Key Formula or Approach:
1. Distance of focus from center: \( c = ae \).
2. Relationship: \( b^2 = a^2(1 - e^2) \).
3. Standard Equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Step 3: Detailed Explanation:
Given foci \( (\pm 3, 0) \), so \( ae = 3 \).
Given \( e = 1/3 \): \[ a(1/3) = 3 \implies a = 9 \implies a^2 = 81 \] Now find \( b^2 \): \[ b^2 = a^2(1 - e^2) = 81(1 - (1/3)^2) = 81(1 - 1/9) \] \[ b^2 = 81(8/9) = 9 \times 8 = 72 \] Substitute into the standard equation: \[ \frac{x^2}{81} + \frac{y^2}{72} = 1 \]
Step 4: Final Answer:
The equation of the ellipse is \( \frac{x^2}{81} + \frac{y^2}{72} = 1 \).