Question

Find the centre, the lengths of the axes, eccentricity, foci of the ellipse:
$\left(ii\right)x^2+4y^2-4x+24y+31=0$

Answer 1

Given: $x^2+4y^2-4x+24y+31=0$

$\Rightarrow (x^2-4x+4)+4(y^2+6y+9)-9=0$

$\Rightarrow (x-2{)}^2+4(y+3{)}^2=9$

$\Rightarrow \frac{(x-2{)}^2}{9}+\frac{(y+3{)}^2}{\frac{9}{4}}=1$

Comparing with $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$

Where, centre $=(h,k)=(2,-3)$

$a^2=9\Rightarrow a=3$

$b^2=\frac{9}{4}\Rightarrow b=\frac{3}{2}$

Length of major axis $=2a=6$

Length of minor axis $=2b=3$

Eccentricity, $e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{\frac{9}{4}}{9}}=\frac{\sqrt{3}}{2}$

Foci $=(h\pm ae,k)=(2\pm \frac{3\sqrt{3}}{2},-3)$

Hence, Centre $=(2,-3)$

Length of Major axis $=6$

Length of Minor axis $=3$

Eccentricity $=\frac{\sqrt{3}}{2}$

Foci $=(2\pm \frac{3\sqrt{3}}{2},-3)$