Question

The eccentricity of the conic 9x² + 25y² = 225 is
(a) 2/5
(b) 4/5
(c) 1/3
(d) 1/5
(e) 3/5

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right. \\ 9x^2+25y^2=225 \\ \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1 \\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}: \\ a=5\mathrm{and}b=3 \\ \mathrm{Here},a>b,\mathrm{so}\mathrm{the}\mathrm{major}\mathrm{and}the\mathrm{minor}\mathrm{ax}es\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{along}thex-axis\mathrm{and}y-\mathrm{axis},\mathrm{respectively}. \\ Now,e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{9}{25}} \\ \Rightarrow e=\sqrt{\frac{16}{25}} \\ \Rightarrow e=\frac{4}{5} \end{gathered}$$