Let $\mathrm{x}=2 \mathrm{t}, \mathrm{y}=\frac{\mathrm{t}^{2}}{3}$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $\mathrm{SA} \perp \mathrm{BA}$, where A is any point on the conic. If k is the ordinate of the centroid of $\Delta \mathrm{SAB}$, then $\lim _{\mathrm{t} \rightarrow 1} \mathrm{k}$ is equal to

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Parabola

Let

x = 2 t, y = \frac{t ^{2}}{3}

be a conic. Let S be the focus and B be the point on the axis of the conic such that

SA ⊥ BA

, where A is any point on the conic. If k is the ordinate of the centroid of

Δ SAB

, then

lim_{t \to 1} k

is equal to

Answer: (D)

From parametric form

x = 2 t, y = \frac{t ^{2}}{3}

, eliminate

t

y = \frac{x ^{2}}{12}

x^{2} = 12 y

.

This is a parabola with focus

S = (0, 3)

and vertex at origin.

Point

A

on parabola:

(2 t, \frac{t ^{2}}{3})

Point

B

is on the axis (y-axis), so

B = (0, α)

.

Given

S A ⊥ B A

m_{S A} \cdot m_{B A} = - 1

\frac{\frac{t ^{2}}{3} - 3}{2 t - 0} \cdot \frac{\frac{t ^{2}}{3} - α}{2 t - 0} = - 1

(\frac{t ^{2}}{3} - 3) (\frac{t ^{2}}{3} - α) = - 4 t^{2}

Solving for

α

3 α = \frac{27 t ^{2} + t ^{4}}{t ^{2} - 9}

Centroid ordinate:

k = \frac{α + \frac{t ^{2}}{3} + 3}{3} = \frac{9 + 3 α + t ^{2}}{9}

lim_{t \to 1} k = lim_{t \to 1} \frac{1}{9} (9 + t^{2} + \frac{27 t ^{2} + t ^{4}}{t ^{2} - 9})

t = 1

k = \frac{1}{9} (10 + \frac{28}{- 8}) = \frac{1}{9} (10 - \frac{7}{2}) = \frac{13}{18}

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option