JEE Main 2022 · 27 July · morning · Single correct · +4 / −1

Let P(a, b) be a point on the parabola y²=8 x such that the tangent at P passes through the centre of the circle x²+y²-10 x-14 y+65=0. Let A be the product of all possible values of a and B be the product of all possible values of b. Then the value of A+B is equal to :

  1. (a) 0
  2. (b) 25
  3. (c) 40
  4. (d) 65
Answer and solution

Answer (d) 65

Solution

Maths sir · 5 steps

Write the tangent at (a, b), make it pass through (5, 7), and you get a quadratic in b.

  1. Centre of circle. For x²+y²+2gx+2fy+c=0 the centre is (-g, -f). Here 2g=-10 and 2f=-14, so the centre is (5, 7).
  2. Tangent at P. For y² = 8x, the tangent at (a, b) is by = 4(x + a). Putting in (5, 7) gives 7b = 20 + 4a.
  3. Use the parabola. P lies on the parabola, so a = b²/8. Then 4a = b²/2, and the condition becomes a quadratic in b.

    7b = 20 + b²/2 ⇒ b² - 14b + 40 = 0 ⇒ b = 4 or 10

  4. Find both products. B = 4 × 10 = 40. The matching values of a are 16/8 = 2 and 100/8 = 12.5, so A = 2 × 12.5 = 25.
  5. So. A + B = 25 + 40 = 65, which is option (d).

The idea to keep. For tangents from an outside point to a parabola, put the point into the tangent at a general point on the curve. You get a quadratic, and Vieta gives products of the coordinates without solving it fully.

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

Study Parabola

Asked about 110 times in JEE Main since 2020. This question is from 2022.