Question

Find the equation of an ellipse with its foci on y-axis, eccentricity 3/4, centre at the origin and passing through (6, 4). 

Answer 1

\[\text{ Let the equation of the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ and } let e=\frac{3}{4}. \text{ Also, let the foci be on they-axis }.\]
\[ \therefore a^2 = b^2 \left( 1 - e^2 \right)\]
\[ \Rightarrow a^2 = b^2 \left( 1 - \frac{9}{16} \right)\]
\[ \Rightarrow a^2 = \frac{7}{16} b^2 \]
\[ \Rightarrow \frac{x^2}{a^2}+\frac{7 y^2}{16 a^2} =1 . . . (1)\]
\[\text{ It passes through }\left( 6, 4 \right).\]
\[\text{ Then } \frac{36}{a^2}+\frac{112}{16 a^2} =1 \]
\[ \Rightarrow \frac{576 + 112}{16 a^2} = 1\]
\[ \Rightarrow 16 a^2 = 688\]
\[ \Rightarrow a^2 = 43\]
\[\text{ Now }, b^2 = \frac{16 a^2}{7}\]
\[ \Rightarrow b^2 = \frac{688}{7}\]
\[\text{ Substituting the values of } a^2 \text{ and } b^2 \text{ in eq }.(1),\text{ we get: }\]
\[\frac{x^2}{43}+\frac{7 y^2}{688} =1 \]
\[\text{ This is the required equation of the ellipse }.\]