Thank you for reporting, we will resolve it shortly
Q.
The foci of the ellipse $25\, x^2 + 4\,y^2 + 100\, x - 4 y + 100 = 0 $ are
TS EAMCET 2017
Solution:
Given equations of ellipse can be rewritten as
$\left((5 \,x)^{2}+2(5)(10) x+10^{2}\right)+$
$\left((2\, y)^{2}-2(2)(1) y+1^{2}\right)-10^{2}-1^{2}+100=0$
$\Rightarrow (5 \,x+10)^{2}+(2 y-1)^{2}=1$
$\Rightarrow 25(x+2)^{2}+4\left(y-\frac{1}{2}\right)^{2}=1$
$\Rightarrow \frac{(x+2)^{2}}{(1 / 5)^{2}}+\frac{(y-1 / 2)^{2}}{(1 / 2)^{2}}=1$, which is of the form
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, 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 $\left(-2, \frac{1}{2} \pm b e\right)=\left(-2, \frac{1}{2} \pm \frac{\sqrt{21}}{10}\right)$
$=\left(-2, \frac{5 \pm \sqrt{21}}{10}\right)$