Q107 / 160 · TS EAMCET 2017 · Mathematics
The foci of the ellipse
$25 x^2+4 y^2+100 x-4 y+100=0$ are
- A$\left(\frac{5 \pm \sqrt{21}}{10},-2\right)$
- Correct answerB$\left(-2, \frac{5 \pm \sqrt{21}}{10}\right)$
- C$\left(\frac{2 \pm \sqrt{21}}{10},-2\right)$
- D$\left(-2, \frac{2 \pm \sqrt{21}}{10}\right)$
Solution
Given equations of ellipse can be rewritten as
$$
\begin{gathered}
\begin{aligned}
&\left((5 x)^2+2(5)(10) x+10^2\right)+ \\
&\left((2 y)^2-2(2)(1) y+1^2\right)-10^2-1^2+100=0 \\
& \Rightarrow \quad(5 x+10)^2+(2 y-1)^2=1
\end{aligned} \\
\Rightarrow \quad 25(x+2)^2+4\left(y-\frac{1}{2}\right)^2=1 \\
\Rightarrow \frac{(x+2)^2}{(1 / 5)^2}+\frac{(y-1 / 2)^2}{(1 / 2)^2}=1, \text { which is of the form } \\
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \text { where } a < b
\end{gathered}
$$
$\Rightarrow \frac{(x+2)^2}{(1 / 5)^2}+\frac{(y-1 / 2)^2}{(1 / 2)^2}=1$, which is of the form
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text {, where } a < b
$$
Here, $a=\frac{1}{5}, b=\frac{1}{2}$ and major axis of ellipse $x+2=0$ i.e. $x=-2$.
Now, $e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{4}{25}}=\sqrt{\frac{21}{25}}=\frac{\sqrt{21}}{5}$
$\therefore$ foci are $\left(-2, \frac{1}{2} \pm\right.$ be $)=\left(-2, \frac{1}{2} \pm \frac{\sqrt{21}}{10}\right)$
$$
=\left(-2, \frac{5 \pm \sqrt{21}}{10}\right)
$$
Hence, the correct answer is option: 'B'.