Question

The eccentricity of the ellipse 4x² + 9y² = 36 is

(a) $\frac{1}{2\sqrt{3}}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\frac{\sqrt{5}}{3}$

(d) $\frac{\sqrt{5}}{6}$

Answer 1

$$\begin{gathered} \left(c\right)\frac{\sqrt{5}}{3} \\ 4x^2+9y^2=36 \\ \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1 \\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}: \\ a=3\mathrm{and}b=2 \\ \mathrm{Here},a>b,\mathrm{so}\mathrm{the}\mathrm{major}\mathrm{and}the\mathrm{minor}\mathrm{ax}es\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{along}thex-axis\mathrm{and}y-\mathrm{axis},\mathrm{respectively}. \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{4}{9}} \\ \Rightarrow e=\sqrt{\frac{5}{9}} \\ \Rightarrow e=\frac{\sqrt{5}}{3} \end{gathered}$$