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

• A

  $2$
• B

  $4$
• C

  $6$
• Correct answer
  D

  $8$

Solution

Solving $\mathrm{y}={\mathrm{x}}^2$ and ${\mathrm{x}}^2+{\left(\mathrm{y}-2\right)}^2=8$
we get, $\mathrm{y}+{\left(\mathrm{y}-2\right)}^2=8$
$\Rightarrow {\mathrm{y}}^2-3\mathrm{y}-4=0\Rightarrow \mathrm{y}=-\mathrm{1,4}$
$\Rightarrow \mathrm{y}=4\mathrm{}$ { $\mathrm{y}\ne -1$ otherwise $\mathrm{x}$ is not defined}
$\Rightarrow {\mathrm{x}}^2=4\Rightarrow \mathrm{x}=\pm 2$
Let, $\mathrm{B}=\left(-\mathrm{2,4}\right), \mathrm{C}=\left(\mathrm{2,4}\right)$

Now, the area of $\mathrm{\Delta }\mathrm{O}\mathrm{B}\mathrm{C}=\frac{1}{2}\times 4\times 4=8$ units

Hence, the correct answer is option: 'D'.