JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Application of Derivatives

The tangents to the curve

y = (x - 2)^{2} - 1

at its points of intersection with the line

x - y = 3

, intersect at the point :

Answer: (D)

Solution

Find where the parabola meets the line first. Writing

y = (x - 2)^{2} - 1 = x^{2} - 4 x + 3

and setting it equal to

y = x - 3

gives

x^{2} - 5 x + 6 = 0

, so

x = 2

and

x = 3

, meeting at

(2, - 1)

and

(3, 0)

. The slope is

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

, which is

0

x = 2

and

2

x = 3

. The tangent at

(2, - 1)

is the horizontal line

y = - 1

, and the tangent at

(3, 0)

y = 2 (x - 3)

. Setting

- 1 = 2 (x - 3)

gives

x = \frac{5}{2}

, so the tangents cross at

(\frac{5}{2}, - 1)

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option