If the tangent at a point P on the parabola $y^{2} = 3x$ is parallel to the line $x + 2y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$, then the area of the triangle PQR is:
(1) $\frac{9}{\sqrt{5}}$
(2) $5\sqrt{3}$
(3) $\frac{3}{2}\sqrt{5}$
(4) $3\sqrt{5}$
Step-by-Step Solution
Key Concept: Find P on parabola where slope $= -1/2$: gives $P(3,-3)$. Tangents on ellipse perpendicular to $x-y=2$ (slope 1) have slope $-1$; find Q and R on ellipse with slope $-1$ tangents. Compute area using determinant formula.
Parabola $y^2 = 3x$: slope at P: $\frac{dy}{dx} = \frac{3}{2y} = -\frac{1}{2} \Rightarrow y_1 = -3$, $x_1 = 3$. So $P(3,-3)$. Ellipse $\frac{x^2}{4}+y^2=1$ with tangents of slope $-1$: $Q\left(\frac{4}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$, $R\left(-\frac{4}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right)$. Area $= \frac{1}{2}\left|3\left(\frac{2}{\sqrt{5}}\right)+3\left(\frac{8}{\sqrt{5}}\right)\right| = \frac{30}{2\sqrt{5}} = 3\sqrt{5}$.
Correct Answer: 4