ExamHabit
Practise free

JEE Main 2019 · 12 Apr · Shift II

JEE Main 2019 Mathematics question on Conic Sections

The tangents to the curve y=(x−2)2−1y = (x-2)^2 - 1 at its points of intersection with the line x−y=3x - y = 3, intersect at the point
  1. A.
    (52,1)\left(\dfrac{5}{2}, 1\right)
  2. B.
    (−52,−1)\left(-\dfrac{5}{2}, -1\right)
  3. C.
    (52,−1)\left(\dfrac{5}{2}, -1\right)
  4. D.
    (−52,1)\left(-\dfrac{5}{2}, 1\right)
Show answer and solution

Correct option: C

Solution

Find intersection of y = (x-2)²-1 and x-y=3, i.e., y=x-3.

(x-2)²-1 = x-3 → x²-4x+4-1 = x-3 → x²-5x+6=0 → x=2,3, giving points (2,-1) and (3,0).

Tangent slope: dy/dx = 2(x-2). At x=2: slope=0, tangent y=-1. At x=3: slope=2, tangent y-0=2(x-3) → y=2x-6.

Intersection of y=-1 and y=2x-6: -1=2x-6 → x=5/2, y=-1. Point is (5/2, -1).

[JEE Main 2019 · 12 Apr · Shift II]

Practise Conic Sections the way it is asked

A timed set of real JEE Main Conic Sections questions, graded as the exam grades. No sign-up.

Start a Conic Sections test

More: all JEE Main Conic Sections questions · JEE Main 2019 · Mathematics chapter list