Question

Find the centre, the lengths of the axes, eccentricity, foci of the ellipse:
$\left(i\right)x^2+2y^2-2x+12y+10=0$

Answer 1

Given: $x^2+2y^2-2x+12y+10=0$

$\Rightarrow (x^2-2x+1)+2(y^2+6y+9)-9=0$

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

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

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

Where, centre $=(h,k)=(1,-3)$

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

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

Length of major axis $=2a=6$

Length of minor axis $=2b=3\sqrt{2}$

Eccentricity,

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

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

Foci $=(h\pm ae,k)=(1\pm \frac{3}{\sqrt{2}},-3)$

Hence,

Centre $=(1,-3)$

Length of Major axis $=6$

Length of Minor axis$=3\sqrt{2}$

Eccentricity$=\frac{1}{\sqrt{2}}$

Foci $=(1\pm \frac{3}{\sqrt{2}},-3)$