ExamHabit
Practise free

JEE Main 2016

JEE Main 2016 Mathematics question on Conic Sections

Let PP be the point on the parabola, y2=8xy^2 = 8x, which is at a minimum distance from the centre CC of the circle, x2+(y+6)2=1x^2 + (y+6)^2 = 1. Then, the equation of the circle, passing through CC and having its centre at PP is
  1. A.
    x2+y2−4x+8y+12=0x^2 + y^2 - 4x + 8y + 12 = 0
  2. B.
    x2+y2−x+4y−12=0x^2 + y^2 - x + 4y - 12 = 0
  3. C.
    x2+y2−x4+2y−24=0x^2 + y^2 - \frac{x}{4} + 2y - 24 = 0
  4. D.
    x2+y2−4x+9y+18=0x^2 + y^2 - 4x + 9y + 18 = 0
Show answer and solution

Correct option: A

Solution

A point on y² = 8x can be written as (2t², 4t). The centre C of the given circle is (0, -6).

Distance² = (2t² - 0)² + (4t + 6)² = 4t⁴ + 16t² + 48t + 36.

Minimize: d/dt(4t⁴ + 16t² + 48t + 36) = 16t³ + 32t + 48 = 0 → t³ + 2t + 3 = 0 → (t+1)(t²-t+3) = 0, so t = -1.

Thus P = (2(1), 4(-1)) = (2, -4).

Radius PC = √((2-0)² + (-4+6)²) = √(4+4) = 2√2.

Circle with centre P(2,-4) and radius 2√2: (x-2)² + (y+4)² = 8 → x² + y² - 4x + 8y + 12 = 0.

[JEE Main 2016]

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 2016 · Mathematics chapter list