Question

Square of the area of the triangle formed by end points of a focal chord PQ of length 32 units of the parabola $y^2 = 8x$ and its vertex, is

• A. 256
• B. 128
• C. 64
• D. 512

Answer 1

Answer: (A)

256

Answer 2

The given parabola is $y^2 = 8x$. Comparing this with the standard form $y^2 = 4ax$, we get $4a = 8$, which implies $a = 2$. The vertex of the parabola is $V(0,0)$. The focus of the parabola is $F(a,0) = F(2,0)$.

Let PQ be a focal chord of length 32 units. The length of a focal chord of a parabola $y^2 = 4ax$ is given by $L = a(t + 1/t)^2$, where $t$ is the parameter of one endpoint. Given $L = 32$ and $a=2$: $32 = 2 (t + 1/t)^2$ $16 = (t + 1/t)^2$ $|t + 1/t| = 4$.

The endpoints of the focal chord are $P(at^2, 2at)$ and $Q(a/t^2, -2a/t)$. The area of the triangle VPQ formed by the vertex $V(0,0)$ and the endpoints P and Q is given by: Area $= \frac{1}{2} |x_P y_Q - x_Q y_P|$ Area $= \frac{1}{2} |(at^2)(-2a/t) - (a/t^2)(2at)|$ Area $= \frac{1}{2} |-2a^2t - 2a^2/t|$ Area $= \frac{1}{2} |-2a^2(t + 1/t)|$ Area $= a^2 |t + 1/t|$

Substituting the values $a=2$ and $|t + 1/t| = 4$: Area $= (2)^2 \times 4 = 4 \times 4 = 16$.

The question asks for the square of the area of the triangle. Square of Area $= (16)^2 = 256$.