Question

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 : 

• A. \(\frac{3}{2}\sqrt 5\)
• B. \(5\sqrt 3\)
• C. \(3\sqrt 5\)
• D. \(\frac{9}{\sqrt 5}\)

Answer 1

The Correct Option is C

(A)The tangent to the parabola \( y^2 = 3x \) has slope \( m \). The equation of the tangent is: \[ y = mx + \frac{1}{m}. \] For parallelism to \( x + 2y = 1 \), \( m = -\frac{1}{2} \). (B)Compute the coordinates of \( P \). Substituting \( y = -\frac{1}{2}x \) into \( y^2 = 3x \): \[ \left(-\frac{1}{2}\right)^2x^2 = 3x \implies x = 4, y = -2. \] (C)For the ellipse, tangents perpendicular to \( x - y = 2 \) have slopes \( m = 1 \). Substituting into tangent conditions, find \( Q \) and \( R \). (D)Use the vertices \( P, Q, R \) to find the area using the determinant formula: \[ \text{Area} = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|. \] Compute to get \( 3\sqrt{5} \).