Question

Find the eccentricity, foci and the length of the latus-rectum of the ellipse $x^2+4y^2+8y-2x+1=0$.

Answer 1

Find eccentricity, foci & length of latus - rectum

of ellipse

$\therefore x^2+4y^2+8y-2x+1=0$

$\therefore x^2-2x+1+4y^2+8y+2-1=0$

$(x-1{)}^2+(2y+2{)}^2-2=0$

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

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

$\therefore \frac{(x-1{)}^2}{2}+\frac{(y+1{)}^2}{1/2}=1$

$\to x_1=1$ & $y_1=-1$

Also,$a=\sqrt{2}$ &$b=\frac{1}{\sqrt{2}}$

centre $=(1,-3)$

$\to$ majer axis$=2a=2\times \sqrt{2}=2\sqrt{2}$

$\to$ minor axis$=2b=2\times \frac{1}{\sqrt{2}}=\sqrt{2}$

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

$e=\sqrt{1-\frac{1/2}{2}}=\sqrt{1-\frac{1}{4}}=\sqrt[0]{0\frac{3}{4}}=\frac{\sqrt{3}}{2}$

$\to$ foci $=(x_1\pm ae,y_1)=(1\pm \sqrt{2}.\frac{\sqrt{3}}{2},-1)$

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

$\to$ length of latus $=\frac{2b^2}{a}=\frac{2\times 1/2}{2}=\frac{1}{2}$ Ans