Question

Find the equation of an ellipse whose axes lie along coordinate axes and which passes through (4, 3) and (−1, 4). 

Answer 1

\[\text{ Let the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ and it passes through the points } (4,3) \text{ and } (-1,4).\]
\[ \therefore \frac{16}{a^2}+\frac{9}{b^2} =1 \text{ and } \frac{1}{a^2}+\frac{16}{b^2} =1 \]
\[\text{ Let } \alpha=\frac{1}{a^2}\text{ and } \beta=\frac{1}{b^2}\]
\[\text{ Then } 16\alpha + 9\beta = 1 \text{ and } \alpha + 16\beta = 1\]
\[\text{ Solving these two equations, we get }:\]
\[\alpha = \frac{7}{247} \text{ and } \beta = \frac{15}{247}\]
\[ \Rightarrow \frac{1}{a^2} = \frac{7}{247}\text{ and } \frac{1}{b^2} = \frac{15}{247} . . . (1) \]
\[\text{Substituting eq. (1) in the equation of an ellipse, we get }:\]
\[\frac{7 x^2}{247} + \frac{15 y^2}{247} = 1\]
\[\text{ This is the required equation of the ellipse }.\]