Question

The foci of the ellipse $25x^2+4y^2+100x-4y+100=0$ are

• A. $\left(\frac{5\pm \sqrt{21}}{10},-2\right)$
• B. $\left(-2,\frac{5\pm \sqrt{21}}{10}\right)$
• C. $\left(\frac{2\pm \sqrt{21}}{10},-2\right)$
• D. $\left(-2,\frac{2\pm \sqrt{21}}{10}\right)$

Answer 1

Answer: (B)

Given equations of ellipse can be rewritten as
$\left((5x{)}^2+2(5)(10)x+10^2\right)+$
$\left((2y{)}^2-2(2)(1)y+1^2\right)-10^2-1^2+100=0$
$\Rightarrow (5x+10{)}^2+(2y-1{)}^2=1$
$\Rightarrow 25(x+2{)}^2+4{\left(y-\frac{1}{2}\right)}^2=1$
$\Rightarrow \frac{(x+2{)}^2}{(1/5{)}^2}+\frac{(y-1/2{)}^2}{(1/2{)}^2}=1$, 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 $\left(-2,\frac{1}{2}\pm be\right)=\left(-2,\frac{1}{2}\pm \frac{\sqrt{21}}{10}\right)$
$=\left(-2,\frac{5\pm \sqrt{21}}{10}\right)$