Question

The eccentricity of the ellipse $25x^2+9y^2-150x-90y+225=0$ is

• A. $\frac{4}{5}$
• B. $\frac{3}{5}$
• C. $\frac{4}{15}$
• D. $\frac{9}{5}$

Answer 1

Answer: (A)

Given curve is $25x^2+9y^2-150x-90y+225=0$
$\Rightarrow$ $(25x^2-150x)+(9y^2-90y)+225=0$
$\Rightarrow$ $25(x^2-6x)+9(y^2-10y)+225=0$
$\Rightarrow$ $25\{x^2-6x+9-9\}+9\{y^2-10y+25-25\}$
$+225=0$
$\Rightarrow$ $25{(x-3)}^2-225+9{(y-5)}^2-225+225=0$
$\Rightarrow$ $25{(x-3)}^2+9{(y-5)}^2=225$
$\Rightarrow$ $\frac{{(x-3)}^2}{9}+\frac{{(y-5)}^2}{25}=1$
$[\because$ $(b>a)$ ]
Which represent an ellipse Whose eccentricity
$e=\sqrt{\frac{b^2-a^2}{b^2}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$