Let quadratic curve passing through the point $(-1,0)$ and touching the line $\mathrm{y}=\mathrm{x}$ at $(1,1)$ be $\mathrm{y}= \mathrm{f}(\mathrm{x})$. Then the x -intercept of the normal to the curve at the point $(\alpha, \alpha+1)$ in the first quadrant is

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Application of Derivatives

Let quadratic curve passing through the point

(- 1, 0)

and touching the line

y = x

(1, 1)

y = f (x)

. Then the x -intercept of the normal to the curve at the point

(α, α + 1)

in the first quadrant is

Answer: 11

Since the curve passes through

(- 1, 0)

, write

f (x) = (x + 1) (a x + b)

. Touching

y = x

(1, 1)

requires

f (1) = 1

and

f^{'} (1) = 1

, giving

a + b = \frac{1}{2}

and

3 a + b = 1

, so

a = b = \frac{1}{4}

and

f (x) = \frac{( x + 1 ) ^{2}}{4}

. For

(α, α + 1)

on the curve,

α + 1 = \frac{( α + 1 ) ^{2}}{4}

gives

α + 1 = 4

, so the first-quadrant point is

(3, 4)

. There

f^{'} (3) = 2

, so the normal is

y - 4 = - \frac{1}{2} (x - 3)

, and setting

y = 0

gives

x = 11

. The x-intercept of the normal is

11

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer