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)

The given ellipse is $25\left(x^2-6x\right)+9\left(y^2-10y\right)=225$
$\Rightarrow 25{\left(x-3\right)}^2+9{\left(y-5\right)}^2$
$=225+225+225=675$
$\Rightarrow \frac{{\left(x-3\right)}^2}{27}+\frac{{\left(y-5\right)}^2}{75}=1$
$\left[\text{Type}\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\right]$
Since $b^2=a^2\left(1-e^2\right)$
$\Rightarrow 1-e^2=\frac{27}{75}$
$\Rightarrow e^2=1-\frac{27}{75}=\frac{48}{75}=\frac{16}{25}$
$\therefore e=\frac{4}{5}$