Question
Question: Point on y<sup>2</sup> = 4x that is nearest to the circle x<sup>2</sup> + (y – 12)<sup>2</sup> = 1, ...
Point on y2 = 4x that is nearest to the circle x2 + (y – 12)2 = 1, is
A
(4, -4)
B
(4, 4)
C
(9, 6)
D
(9, -6)
Answer
(4, 4)
Explanation
Solution
Shortest distance between two curves always occur along the common normal. That means if 'P' is the required point then PA becomes the common normal.
Let P ≡ (t2, 2t). Equation of normal at 'P' is :
y = −tx + 2t + t3, it should pass through (0, 12).
Thus t3 + 2t −12 = 0
⇒ t = 2. Hence P = (4, 4).