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 $
$\quad \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} $