JEE MainMathematicsParabola
The minimum distance between a point on the parabola y^2 = 4x and a point on the circle x^2 + y^2 - 12x + 32 = 0 is
Options
- A2 5
- B4
- C2 5 + 2
- D2 5 - 2
Correct answer
D. 2 5 - 2
Step-by-step solution
The shortest distance between two non-intersecting curves lies along their common normal. For a circle, any normal must pass through its centre. The given circle is x^2 + y^2 - 12x + 32 = 0 . Rewriting it in standard form: (x - 6)^2 + y^2 = 4 . The centre of the circle is C(6, 0) and its radius is r = 2 . The equation of the parabola is y^2 = 4x , which means a = 1 . The parametric equation of the normal to the parabola y^2 = 4ax at point (at^2, 2at) is y = -tx + 2at + at^3 . For a = 1 , the normal is y = -tx + 2t