Question

Find the centre, the lengths of the axes, eccentricity, foci of the ellipse:
$\left(iii\right)4x^2+y^2-8x+2y+1=0$

Answer 1

Given: $4x^2+y^2-8x+2y+1=0$

$\Rightarrow 4(x^2-2x+1)+(y^2+2y+1)-4=0$

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

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

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

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

$a^2=1\Rightarrow a=1$

$b^2=4\Rightarrow b=2$

Length of minor axis $=2a=2$ $(\because a<b)$

Length of major axis $=2b=4$

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

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

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

Hence,

Centre $=(1,-1)$

Length of Major axis $=4$

Length of Minor axis $=2$

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

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