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 $x\ne -1$ is rejected because it makes $y$ not defined)
$\Rightarrow y^2=4\Rightarrow y=\pm 2$
Let, $B=\left(4,-2\right),C=\left(4,2\right)$

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