Length of latus rectum of the ellipse $2 x^{2} + y^{2} - 8 x + 2 y + 7 = 0$ is

\sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\sqrt{2}$
Ellipse: $2 x^{2} + y^{2} - 8 x + 2 y + 7 = 0$
$\Rightarrow 2 (x^{2} - 4 x) + y^{2} + 2 y + 7 = 0$
$\Rightarrow 2 (x^{2} - 4 x + 4) - 2 (4) + (y + 2 y + 1) + 7 - 1 = 0$
$\Rightarrow 2 (x - 2)^{2} + (y + 1)^{2} = 2$
$\Rightarrow \frac{(x - 2)^{2}}{1} + \frac{(y + 1)^{2}}{2} = 1$
here $a^{2} = 1$ and $b^{2} = 2$
so, $b > a$
length of $L . R$ = $2 \frac{a^{2}}{b} = \frac{2 (1)}{\sqrt{2}} = \sqrt{2}$

Suggest Corrections

Similar questions

2 x^{2} + y^{2} - 8 x - 2 y + 1 = 0.

y^{2} + 8 x - 2 y + 17 = 0

4 x^{2} + 9 y^{2} + 8 x + 36 y + 4 = 0

The length of the latus-rectum of the parabola $y^{2} + 8 x - 2 y + 17 = 0$ is

Join BYJU'S Learning Program