A) \[\sqrt{2}\]
B) \[2\]
C) \[8\]
D) None of these
Correct Answer: A
Solution :
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\[=\frac{2{{a}^{2}}}{b}\] \[=2\cdot \frac{1}{\sqrt{2}}=\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
