Discussion Forum

The foci of the ellipse

$25x^{2}+4y^{2}+100x-4y+100=0$ are

Answer:

Explanation:

 Given equation of ellipse can be rewritten as 

$((5x)^{2}+2(5)(10)x+10^{2})+$

                   $((2y)^{2}-2(2)(1)y+1^{2})-10^{2}-1^{2}+100=0$

$\Rightarrow$    (5x+10)^{2}+(2y-1)^{2}=1$

$\Rightarrow$  $25(x+2)^{2}+4(y-\frac{1}{2})^{2}=1$

$\Rightarrow$    $\frac{(x+2)^{2}}{(1/5)^{2}}+\frac{(y-\frac{1}{2})^{2}}{(1/2)^{2}}=1$ , 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   $(-2,\frac{1}{2}\pm be)=\left(-2,\frac{1}{2}\pm \frac{\sqrt{21}}{10}\right)$

    $=\left(-2,\frac{5\pm\sqrt{21}}{10}\right)$