Question

Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point (−3, 1) and has eccentricity equal to .

Answer 1

$$\begin{gathered} \text{Let the equation of the ellipse be}\frac{x^2}{a^2}\text{+}\frac{y^2}{b^2}\text{=1 ...(1)} \\ \text{It passes through the point}\left(-3,1\right)\text{.} \\ \text{∴}\frac{9}{a^2}\text{+}\frac{1}{b^2}\text{=1 ...( 2)} \\ \mathrm{and}e=\sqrt{\frac{2}{5}} \\ \mathrm{Now},b^2=a^2\left(1-e^2\right) \\ \Rightarrow b^2=a^2\left(1-\frac{2}{5}\right) \\ \Rightarrow b^2=a^2\times \frac{3}{5}\mathrm{or}\frac{3a^2}{5} \\ \text{Substituting the value of}{\mathit{\text{b}}}^2\text{in eq. (2), we get:} \\ \frac{9}{a^2}\text{+}\frac{5}{3a^2}\text{=1} \\ \Rightarrow \frac{27+5}{3a^2}=1 \\ \Rightarrow a^2=\frac{32}{3} \\ \Rightarrow b^2=\frac{3\times {\displaystyle \frac{32}{3}}}{5}\mathrm{or}\frac{32}{5} \\ \text{Substituting the values of}\mathit{\text{a}}\text{and}\mathit{\text{b}}\text{in eq. (1), we get:} \\ \frac{3x^2}{32}+\frac{5y^2}{32}=1 \\ \Rightarrow 3x^2+5y^2=32 \\ \mathrm{This}\text{is the required equation of the ellipse.} \end{gathered}$$