Question

The length of latusrectum of the ellipse $ 2{{x}^{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 $ 2{{x}^{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{2{{a}^{2}}}{b} $
$ =2\cdot \frac{1}{\sqrt{2}}=\sqrt{2} $