Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $\mathrm{A}_{1}$ is twice the area $\mathrm{A}_{2}$. Then the normal to the curve perpendicular to the line $2 \mathrm{x}-12 \mathrm{y}=15$ does NOT pass through the point.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Application of Derivatives

Consider a curve

y = y (x)

in the first quadrant as shown in the figure. Let the area

A_{1}

is twice the area

A_{2}

. Then the normal to the curve perpendicular to the line

2 x - 12 y = 15

does NOT pass through the point.

Answer: (C)

From the figure,

A_{1} + A_{2} = x y - 8

. Given

A_{1} = 2 A_{2}

, we have

\frac{3 A _{1}}{2} = x y - 8

, so

A_{1} = \frac{2 x y - 16}{3}

.

Since

A_{1} = \int_{4}^{x} y (t) d t

, differentiating:

y = \frac{d}{d x} (\frac{2 x y - 16}{3}) = \frac{2}{3} (x \frac{d y}{d x} + y)

.

This gives

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

, leading to

y^{2} = c x

. Using

f (4) = 2

, we get

c = 1

, so

y^{2} = x

.

The line

2 x - 12 y = 15

has slope

\frac{1}{6}

. A normal perpendicular to it has slope

- 6

.

For the curve

y^{2} = x

\frac{d y}{d x} = \frac{1}{2 y}

. The normal has slope

- 2 y

. Setting

- 2 y = - 6

gives

y = 3

, so

x = 9

.

The normal at

(9, 3)

is:

y - 3 = - 6 (x - 9)

, i.e.,

y + 6 x = 57

.

Checking the options:

(10, - 4)

gives

- 4 + 60 = 56 \neq = 57

, so it does NOT lie on the normal.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option