Question

The eccentricity of the ellipse 25x² + 16y² = 400 is
(a) 3/5
(b) 1/3
(c) 2/5
(d) 1/5

Answer 1

$$\begin{gathered} \left(a\right)\frac{3}{5} \\ 25x^2+16y^2=400 \\ \Rightarrow \frac{x^2}{16}+\frac{y^2}{25}=1...\left(1\right) \\ \mathrm{Comparing}\mathrm{equation}\left(1\right)\mathrm{with}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}: \\ {a}^2=16\mathrm{and}{b}^2=25 \\ \mathrm{Here},a<b,\mathrm{so}\mathrm{the}\mathrm{major}\mathrm{and}\mathrm{the}\mathrm{minor}\mathrm{axes}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{ellipse}\mathrm{are}\mathrm{along}\mathrm{the}y-axis\mathrm{and}\mathrm{the}x-\mathrm{axis},\mathrm{respectively}. \\ \mathrm{Now},e=\sqrt{1-\frac{a^2}{b^2}} \\ \Rightarrow e=\sqrt{1-\frac{16}{25}} \\ \Rightarrow e=\sqrt{\frac{9}{25}} \\ \Rightarrow e=\frac{3}{5} \end{gathered}$$