MCQ
The eccentricity of the conic $9\text{x}^2+25\text{y}^2=225$ is:
  • A
    $\frac{2}{5}$
  • $\frac{4}{5}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{5}$

Answer

Correct option: B.
$\frac{4}{5}$
$\Rightarrow\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
Comparing it with $\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ we get:
$\text{a}=5$ and $\text{b}=3$
Here, $a > b,$ so the major and the minor axes of the ellipse are along the $x−$axis and $y−$axis, respectively.
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{9}{25}}$
$\Rightarrow\text{e}=\sqrt{\frac{16}{25}}$
$\Rightarrow\text{e}=\frac{4}{5}$

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