Let the tangent to the parabola $\mathrm{S}: \mathrm{y}^{2}=2 \mathrm{x}$ at the point $\mathrm{P}(2,2)$ meet the x -axis at Q and normal at it meet the parabola S at the point R . Then the area (in sq. units) of the triangle PQR is equal to :

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Parabola

Let the tangent to the parabola

S : y^{2} = 2 x

at the point

P (2, 2)

meet the x -axis at Q and normal at it meet the parabola S at the point R . Then the area (in sq. units) of the triangle PQR is equal to :

Answer: (D)

P (2, 2)

the parabola

y^{2} = 2 x

has tangent

y y_{1} = x + x_{1}

, that is

2 y = x + 2

, which cuts the

x

-axis at

Q (- 2, 0)

. The tangent has slope

\frac{1}{2}

, so the normal at

P

has slope

- 2

y - 2 = - 2 (x - 2) \Rightarrow y = 6 - 2 x

Putting

x = \frac{y ^{2}}{2}

into this gives

y^{2} + y - 6 = 0

, whose root other than

y = 2

y = - 3

, so the normal meets the parabola again at

R (\frac{9}{2}, - 3)

. For the vertices

P (2, 2)

Q (- 2, 0)

and

R (\frac{9}{2}, - 3)

Area = \frac{1}{2} 2 (0 + 3) + (- 2) (- 3 - 2) + \frac{9}{2} (2 - 0) = \frac{1}{2} (6 + 10 + 9) = \frac{25}{2}

so triangle

P QR

encloses

\frac{25}{2}

sq. units.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option