Question

The length of latusrectum of the ellipse $2x^2+y^2-8x+2y+7=0$ , is

• A. $\sqrt{2}$
• B. $2$
• C. $8$
• D. None of these

Answer 1

Answer: (A)

We have, the ellipse $2x^2+y^2-8x+2y+7=0$
$2(x^2-4x)+(y^2+2y)+7=0$
$\Rightarrow$ $2{(x-2)}^2-8+{(y+1)}^2-1+7=0$
$\Rightarrow$ $2{(x-2)}^2+{(y+1)}^2=2$
$\Rightarrow$ $\frac{{(x-2)}^2}{1}+\frac{{(y+1)}^2}{2}=1$
Comparing this equation with
$\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$ ,
we get $a^2=1,b^2=2$
$\therefore$ Length of latusrectum $=\frac{2a^2}{b}$
$=2\cdot \frac{1}{\sqrt{2}}=\sqrt{2}$