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)

$\frac{4}{5}$

Answer 2

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 $b^{2} = a^{2} \left(1-e^{2} \right)$ $\Rightarrow 1-e^{2} = \frac{27}{75}$ $\Rightarrow e^{2}= 1-\frac{27}{75} = \frac{48}{75} = \frac{16}{25}$ $\therefore e=\frac{4}{5}$