Question

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

Answer 1

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