Question

The eccentricity of the ellipse 4x² + 9y² + 8x + 36y + 4 = 0 is

(a) $\frac{5}{6}$

(b) $\frac{3}{5}$

(c) $\frac{\sqrt{2}}{3}$

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

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)\frac{\sqrt{5}}{3} \\ 4x^2+9y^2+8x+36y+4=0 \\ \Rightarrow 4(x^2+2x)+9(y^2+4y)=-4 \\ \Rightarrow 4(x^2+2x+1)+9(y^2+4y+4)=-4+4+36 \\ \Rightarrow 4(x+1{)}^2+9(y+2{)}^2=36 \\ \Rightarrow \frac{4(x+1{)}^2}{36}+\frac{9(y+2{)}^2}{36}=1 \\ \Rightarrow \frac{(x+1{)}^2}{9}+\frac{(y+2{)}^2}{4}=1 \\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}: \\ a=3andb=2 \\ So,\mathrm{the}\mathrm{major}\mathrm{and}the\mathrm{minor}\mathrm{ax}es\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{along}thex-axis\mathrm{and}y-\mathrm{axis},\mathrm{respectively}. \\ Now,e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{4}{9}} \\ \Rightarrow e=\sqrt{\frac{5}{9}} \\ \Rightarrow e=\frac{\sqrt{5}}{3} \end{gathered}$$