Question

If a chord of the parabola $y^2=4x$ subtend a right angle at the centre of the circle $x^2+y^2=16$ and the chord passes through a fixed point P and from point P three normal are drawn to intersect the parabola at A, B, C. Then Area of $\triangle ABC$ is

• A. 8\sqrt{2}
• B. 4\sqrt{2}
• C. 2\sqrt{2}
• D. 6\sqrt{2}

Answer 1

Answer: (B)

4\sqrt{2}

Answer 2

The condition that a chord of $y^2=4x$ subtends a right angle at the origin implies $t_1t_2 = -4$ for its endpoints' parameters. The equation of such a chord is $2x - y(t_1+t_2) - 8 = 0$. For this chord to pass through a fixed point P for all possible $t_1$, P must be $(4,0)$. Normals from $P(4,0)$ to $y^2=4x$ have parameters $t$ satisfying $t^3 - 2t = 0$, yielding $t=0, \sqrt{2}, -\sqrt{2}$. These correspond to points $A(0,0)$, $B(2, 2\sqrt{2})$, and $C(2, -2\sqrt{2})$. The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base BC} \times \text{height} = \frac{1}{2} \times (4\sqrt{2}) \times 2 = 4\sqrt{2}$.