Q89 / 180 · COMEDK 2013 · Mathematics

For the ellipse $25 x^{2}+9 y^{2}-150 x-90 y+225=0$ the eccentricity $e$ is equal to

  • A
    $\frac{2}{5}$
  • B
    $\frac{3}{5}$
  • Correct answer
    C
    $\frac{4}{5}$
  • D
    $\frac{1}{5}$

Solution

We have, $\begin{aligned} & 25 x^{2}+9 y^{2}-150 x-90 y+225=0 \\ \Rightarrow & 25\left(x^{2}-6 x\right)+9\left(y^{2}-10 y\right)+225=0 \\ \Rightarrow & 25\left[(x-3)^{2}-9\right]+9\left[(y-5)^{2}-25\right]+225=0 \\ \Rightarrow & 25(x-3)^{2}-225+9(y-5)^{2}-225+225=0 \\ \Rightarrow & 25(x-3)^{2}+9(y-5)^{2}=225 \\ \Rightarrow & \quad a=3, b=5 \\ \text { On comparing with } \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1, \text { we have } \\ \therefore & \quad \text { Eccentricity, } e=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \end{aligned}$ $\therefore$ Eccentricity, $e=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$

Hence, the correct answer is option: 'C'.

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