Question

The foci of an ellipse are (±3, 0) and its eccentricity is 1/3, find its equation.
1. \(\frac{x^2}{81}+\frac{y^2}{72}=1\)
2. \(\frac{x^2}{9}+\frac{y^2}{8}=1\)
3. \(\frac{x^2}{8}+\frac{y^2}{9}=1\)
4. \(\frac{x^2}{3}+\frac{y^2}{2}=1\)

Answer 1

Correct Answer - Option 1 : \(\frac{x^2}{81}+\frac{y^2}{72}=1\)

Concept:

The general equation of the ellipse is:

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Here, coordinates of foci are (±ae, 0).

Also, we have b² = a²(1 - e²), where e is the eccentricity.

Calculation:

Since the coordinates of the foci are (±3, 0).

⇒ ae = 3

⇒ a × (1/3) = 3      (∵ e = 1/3)

⇒ a = 9

Now, b2 = a2(1 - e2)

\(⇒ b^{2}=81\left ( 1-\frac{1}{9} \right )\)

⇒ b² = 72

On putting the value of a² and b² in the general equation of an ellipse, we get

\(\frac{x^2}{81}+\frac{y^2}{72}=1\)

Hence, the equation of the ellipse is \(\frac{x^2}{81}+\frac{y^2}{72}=1\).