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:

For ellipse with foci \((\pm c,0)\), use \(e=\frac{c}{a}\) and \(b^2=a^2-c^2\).

Answer 1

The Correct Option is A

Concept: For an ellipse with major axis along \(x\)-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] where \[ e=\frac{c}{a} \] and \[ b^2=a^2-c^2 \]

Step 1: Given foci are: \[ (\pm 3,0) \] So: \[ c=3 \]

Step 2: Given eccentricity: \[ e=\frac{1}{3} \] Using: \[ e=\frac{c}{a} \] \[ \frac{1}{3}=\frac{3}{a} \] \[ a=9 \] So: \[ a^2=81 \]

Step 3: Find \(b^2\). \[ b^2=a^2-c^2 \] \[ b^2=81-9 \] \[ b^2=72 \]

Step 4: Write the equation of ellipse. \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] \[ \frac{x^2}{81}+\frac{y^2}{72}=1 \] Therefore, \[ \boxed{\frac{x^2}{81}+\frac{y^2}{72}=1} \]