Question

Find the equation of an ellipse whose foci are at (± 3, 0) and which passes through (4, 1).

Answer 1

\[\text{ Let the equation of the ellipse be } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 . . . (1)\]
\[\text{ Then ae } = 3\]
\[ \text{ Also } x = 4 \text{ and } y = 1 [\text{ Ellipse passing through } (4, 1)]\]
\[\text{ Substituting the values of x and y in eq  } . (1), \text{ we get }: \]
\[\frac{4^2}{a^2} + \frac{1^2}{b^2} = 1\]
\[ \Rightarrow \frac{16}{a^2} + \frac{1}{b^2} = 1\]
\[\text{ Now }, b^2 = a^2 (1 - e^2 )\]
\[ \Rightarrow b^2 = a^2 - a^2 e^2 \]
\[ \Rightarrow b^2 = a^2 - 9 \text{ or } a^2 = b^2 + 9 . . . \left( 2 \right)\]
\[ \Rightarrow \frac{16}{a^2} + \frac{1}{b^2} = 1\]
\[ \Rightarrow 16 b^2 + a^2 = a^2 b^2 \]
\[ \Rightarrow 16 b^2 + b^2 + 9 = b^2 \left( b^2 + 9 \right)\]
\[ \Rightarrow 17 b^2 + 9 = b^4 + 9 b^2 \]
\[ \Rightarrow b^4 - 8 b^2 - 9 = 0\]
\[ \Rightarrow \left( b^2 - 9 \right)\left( b^2 + 1 \right)\]
\[ \Rightarrow b = \pm 3\]
\[\text{ Substituting the value ofbin eq. } (2), \text{ we get }:\]
\[a = 3\sqrt{2}\]
\[ \therefore \frac{x^2}{18} + \frac{y^2}{9} = 1\]
\[\text{ This is the required equation of the ellipse }.\]