Question

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

Answer 1

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

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

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

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

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

Where, centre $=(h,k)=(1,0)$

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

$b^2=\frac{1}{4}\Rightarrow b=\frac{1}{2}$

Length of major axis $=2a=2$

Length of minor axis $=2b=1$

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

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

Foci $=(h\pm ae,k)=(1\pm \frac{\sqrt{3}}{2},0)$

Hence,

Centre $=(1,0)$

Length of Major axis $=2$

Length of Minor axis $=1$

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

Foci $=(1\pm \frac{\sqrt{3}}{2},0)$