Question

For the following ellipse, find the length of major and minor axes, coordinates of foci and vertices and the eccentricity.

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

Answer 1

$\because x^2+4y^2-2x=0$

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

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

$\Rightarrow \frac{{(x-1)}^2}{1}+\frac{y^2}{\frac{1}{4}}=1\Rightarrow \frac{{(x-1)}^2}{1^2}+\frac{(y-0{)}^2}{{\left(\frac{1}{2}\right)}^2}=1oncomparingwith:\frac{(x-x_0{)}^2}{a^2}+\frac{{(y-y_0)}^2}{b^2}=1weget,a=1andb=\frac{1}{2}alsothecentreofellipseisin(1,0)$

hence, major axis is along x-axis

therefore length of major axis is $=2a=2\times 1=2$

and length of minor axis is $=2b=2\times \frac{1}{2}=1$

$\therefore vertices=(x_0\pm a,y_0)=(1\pm 1,0)=(2,0);(0,0)$

$\because c^2=a^2-b^2=1-\frac{1}{4}=\frac{3}{4}\Rightarrow c=\frac{\sqrt{3}}{2}\therefore foci=(x_0\pm c,y_0)=(1\pm \frac{\sqrt{3}}{2},0)$

$andeccentricity=\frac{c}{a}=\frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}$