Question

Point $A$ is the vertex of the parabola whose equation is $y=x^2-2$. Points $B$ and $C$ are the intersections of the parabola with the circle whose equation is $x^2+y^2=8$. The number of square units in the area of $△ABC$, is

• A. 8
• B. 4
• C. 12
• D. 16

Answer 1

Answer: (A)

$y=x^2-2;x^2+y^2=8$
$x^2=y+2;y^2+y+2=8$
$y^2+y-6=0\Rightarrow y^2+3y-2y-6=0$
$\Rightarrow (y+3)(y-2)=0\Rightarrow y=2\text{ or }y=-3\text{ (rejected) }$
$\text{ If }y=2;x=2\text{ or }-2$
$\therefore B(2,2);C=(-2,2)\text{ and }A(0,-2)$
$\Rightarrow \text{ Area of }△ABC=\frac{4\times 4}{2}=8$