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JEE Main Mathematics question on Conic Sections

The tangent to the parabola y2=4xy^2 = 4x at the point where it intersects the circle x2+y2=5x^2 + y^2 = 5 in the first quadrant, passes through the point
  1. A.
    (14,34)\left(\dfrac{1}{4}, \dfrac{3}{4}\right)
  2. B.
    (34,74)\left(\dfrac{3}{4}, \dfrac{7}{4}\right)
  3. C.
    (−13,23)\left(-\dfrac{1}{3}, \dfrac{2}{3}\right)
  4. D.
    (−14,12)\left(-\dfrac{1}{4}, \dfrac{1}{2}\right)
Show answer and solution

Correct option: B

Solution

The parabola y² = 4x can be parametrized as (t², 2t). Substituting into x² + y² = 5: t⁴ + 4t² - 5 = 0, giving (t² - 1)(t² + 5) = 0, so t = 1 (first quadrant). The intersection point is (1, 2).

The tangent to y² = 4x at (1, 2): differentiating, 2y·dy/dx = 4, so dy/dx = 2/y = 1 at (1,2). Tangent line: y - 2 = 1(x - 1), i.e., y = x + 1.

Check option 2: x = 3/4, y = 3/4 + 1 = 7/4. ✓

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