If P(h, k) be a point on the parabola x=4 y², which is nearest to the point Q(0,33), then the distance of P from the directrix of the parabola y²=4(x+y) is equal to :

Solution

Maths sir · 4 steps

Minimise the squared distance to Q to find P, then complete the square to find the directrix.

1. Put P on curve. Any point on x = 4y² is (4k², k). Its squared distance from Q(0, 33) is D = 16k⁴ + (k - 33)². You do not need the square root, because the smallest D also gives the smallest distance.

   dD/dk = 64k³ + 2(k - 33) = 0 ⇒ 32k³ + k - 33 = 0

2. Find the nearest point. k = 1 works, since 32 + 1 - 33 = 0. Dividing out (k - 1) leaves 32k² + 32k + 33. Its discriminant 32² - 4 × 32 × 33 is negative, so k = 1 is the only critical point, D is smallest there, and P = (h, k) = (4, 1).
3. Locate the directrix. Rewrite y² = 4x + 4y as y² - 4y + 4 = 4x + 4, that is (y - 2)² = 4(x + 1). The vertex is (-1, 2) and a = 1, so the directrix is x + 1 = -1, which is x = -2.
4. So. The distance of P(4, 1) from the vertical line x = -2 is 4 - (-2) = 6, which is option (c).

The idea to keep. For the nearest point on a curve, minimise the squared distance in terms of the curve's parameter. For a shifted parabola, complete the square first, then read off the vertex and a.

Answer from the NTA final answer key. Solution by PrepWiser.