Question

Let $B$ and $C$ are the points of intersection of the parabola $x=y^{2}$ and the circle $y^{2}+\left(x - 2\right)^{2}=8.$ The perimeter (in units) of the triangle $OBC,$ where $O$ is the origin, is

• A. $8$
• B. $4\sqrt{5}$
• C. $4\sqrt{5}+2$
• D. $4\left(\sqrt{5} + 1\right)$

Answer 1

Answer: (D)

On solving $x=y^{2}$ and $y^{2}+\left(x - 2\right)^{2}=8$
we get, $x+\left(x - 2\right)^{2}=8$
$\Rightarrow x^{2}-3x-4=0\Rightarrow x=-1,4$
$\Rightarrow x=4$ (as $x\neq -1$ is rejected because it makes $y$ not defined)
$\Rightarrow y^{2}=4\Rightarrow y=\pm2$
Let, $B=\left(4 , - 2\right),C=\left(4 , 2\right)$

Now, perimeter of $\Delta OBC=4+4\sqrt{5}$ units