Let A be the vertex of parabola y = x 2-2 and B , C are the intersection points of the parabola with the circle x2+y2=8. Then the area of triangle A B C is

Question

Let A be the vertex of parabola y = x 2-2 and B , C are the intersection points of the parabola with the circle x2+y2=8. Then the area of triangle A B C is (A) 8 (B) 9 (C) 10 (D) 12

Tardigrade Admin · Accepted Answer

Solve P S B ≡(2,2) ∴ (-2,2) ∴ A=(1/2) × 4 × 4=8

Q. Let $A$ be the vertex of parabola $y = x^{2} - 2$ and $B, C$ are the intersection points of the parabola with the circle $x^{2} + y^{2} = 8$ . Then the area of triangle $A BC$ is

8

9

10

12

Solve P & S
$B \equiv (2, 2)$
$∴ (- 2, 2)$
$∴ A = \frac{1}{2} \times 4 \times 4 = 8$

Q. Let A be the vertex of parabola y=x2−2 and B,C are the intersection points of the parabola with the circle x2+y2=8. Then the area of triangle ABC is

8

9

10

12

Solve P & S B≡(2,2) ∴(−2,2) ∴A=21​×4×4=8

Q. Let $A$ be the vertex of parabola $y = x^{2} - 2$ and $B, C$ are the intersection points of the parabola with the circle $x^{2} + y^{2} = 8$ . Then the area of triangle $A BC$ is

Solve P & S
$B \equiv (2, 2)$
$∴ (- 2, 2)$
$∴ A = \frac{1}{2} \times 4 \times 4 = 8$