JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Ellipse

If the tangent at a point P on the parabola

y^{2} = 3 x

is parallel to the line

x + 2 y = 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:

Answer: (D)

Solution

y^{2} = 3 x

\frac{d y}{d x} = \frac{3}{2 y}

; setting this equal to the slope

- \frac{1}{2}

x + 2 y = 1

gives

y = - 3

and

x = 3

, so

P = (3, - 3)

. For the ellipse

\frac{x ^{2}}{4} + y^{2} = 1

, a tangent perpendicular to

x - y = 2

has slope

- 1

; at the point of tangency

x_{0} = 4 y_{0}

, and

\frac{x _{0}^{2}}{4} + y_{0}^{2} = 1

gives

Q = (\frac{4}{5}, \frac{1}{5})

and

R = (- \frac{4}{5}, - \frac{1}{5})

. The area of

△ P QR

\frac{1}{2} 3 \cdot \frac{2}{5} + \frac{4}{5} (3 - \frac{1}{5}) + (- \frac{4}{5}) (- 3 - \frac{1}{5}) = \frac{1}{2} \cdot \frac{30}{5} = 35

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option