Question

Find the equation of an ellipse whose vertices are (0, ± 10) and eccentricity e = $\frac{4}{5}$.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{ellipse}\mathrm{be} \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1...\left(1\right) \\ \mathrm{Since}\mathrm{the}\mathrm{vertices}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{on}\mathrm{the}y-\mathrm{axis},\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{vertices}\mathrm{are}(0,\pm b). \\ \therefore b=10 \\ \mathrm{Now},a^2=b^2(1-e^2) \\ \Rightarrow a^2=100{\left(1-\frac{4}{5}\right)}^2 \\ \Rightarrow a^2=100\times \left(\frac{9}{25}\right) \\ \Rightarrow a^2=36 \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}{a}^2\mathrm{and}{b}^2\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ \frac{x^2}{36}+\frac{y^2}{100}=1 \\ \Rightarrow \frac{100x^2+36y^2}{3600}=1 \\ \therefore 100x^2+36y^2=3600 \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{ellipse}. \end{gathered}$$