Let $PQ$ be a chord of the parabola $y^2 = 4x$. A circle drawn with $PQ$ as a diameter passes through the vertex $V$ of the parabola. If area $(APQ) = 20 \text{ unit}^2$ then the coordinates of $P$ is/are
Step-by-Step Solution
Key Concept: A circle with chord PQ as diameter passes through vertex V if and only if ∠PVQ = 90°, meaning VP ⊥ VQ. For a parabola y² = 4x with parametric points P(t²,2t) and Q(s²,2s), the perpendicularity condition ts = -4 must hold, combined with the area constraint ½|t-s|·t²·|s| = 20 to find t.
The slope of $PV$ is $m = \frac{2t - 0}{t^2 - 0} = \frac{2}{t}$, so the equation of line $QV$ perpendicular to it is $y = -\frac{t}{2}x$. Solving this with the parabola $y^2 = 4x$ gives $Q = \left(\frac{16}{t^2}, \frac{-8}{t}\right)$. Using the area formula $\text{ar}(\triangle PVQ) = \frac{1}{2} \times PV \times VQ = 20$, we get $PV^2 \cdot VQ^2 = 40^2$, which simplifies to $(t^2 - 16)(t^2 - 1) = 0$, giving $t = \pm 4, \pm 1$.
Correct Answer: 1,2