Question

The eccentricity of the ellipse 9x² + 25y² − 18x − 100y − 116 = 0, is
(a) 25/16
(b) 4/5
(c) 16/25
(d) 5/4

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\frac{4}{5} \\ 9x^2-18x+25y^2-100y-116=0 \\ \Rightarrow 9(x^2-2x)+25(y^2-4y)=116 \\ \Rightarrow 9(x^2-2x+1)+25(y^2-4y+4)=116+100+9 \\ \Rightarrow 9(x-1{)}^2+25(y-2{)}^2=225 \\ \Rightarrow \frac{9(x-1{)}^2}{225}+\frac{25(y-2{)}^2}{225}=1 \\ \Rightarrow \frac{(x-1{)}^2}{25}+\frac{(y-2{)}^2}{9}=1 \\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}: \\ a=5\mathrm{and}b=3 \\ \mathrm{Here},a>b,\mathrm{so}\mathrm{the}\mathrm{major}\mathrm{and}\mathrm{the}\mathrm{minor}\mathrm{ax}es\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{along}thex-axis\mathrm{and}y-\mathrm{axis},\mathrm{respectively}. \\ Now,e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{9}{25}} \\ \Rightarrow e=\sqrt{\frac{16}{25}} \\ \Rightarrow e=\frac{4}{5} \end{gathered}$$