Question

Find the equation of the ellipse whose foci are (4, 0) and (−4, 0), eccentricity = 1/3. 

Answer 1

\[\text{ Let the equation of the ellipse be } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 . . . (1)\]
\[\text{ Then the coordinates of the foci are } ( \pm ae, 0) . \]
\[\text{ Also }, e = \frac{1}{3}\]
\[\text{ We have } ae = 4\]
\[ \Rightarrow a . \frac{1}{3} = 4\]
\[ \Rightarrow a = 12\]
\[\text{ Now }, b^2 = a^2 (1 - e^2 )\]
\[ \Rightarrow b^2 = {12}^2 \left( 1 - \left( \frac{1}{3} \right) \right)^2 \]
\[ \Rightarrow b^2 = 144\left( \frac{8}{9} \right)\]
\[ \Rightarrow b^2 = 128\]
\[\text{ Substituting the values of } a^2 \text{ and } b^2 \text{ in eq } . (1), \text{ we get }: \]
\[\frac{x^2}{144} + \frac{y^2}{128} = 1\]
\[\text{ This is the required equation of the ellipse } .\]