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JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Mathematics question on Conic Sections

The equation of a tangent to the parabola, x2=8yx^2 = 8y, which makes an angle θ\theta with the positive direction of XX-axis, is
  1. A.
    y=xtan⁡θ−2cot⁡θy = x\tan\theta - 2\cot\theta
  2. B.
    x=ycot⁡θ+2tan⁡θx = y\cot\theta + 2\tan\theta
  3. C.
    y=xtan⁡θ+2cot⁡θy = x\tan\theta + 2\cot\theta
  4. D.
    x=ycot⁡θ−2tan⁡θx = y\cot\theta - 2\tan\theta
Show answer and solution

Correct option: B

Solution

For x² = 8y, we have 4a = 8, so a = 2. The tangent to x² = 4ay with slope m is y = mx - am². Here m = tanθ and a = 2, giving y = x tanθ - 2tan²θ.

Rearranging: x = y cotθ + 2tan²θ · cotθ = y cotθ + 2tanθ.

This matches option 2: x = y cotθ + 2tanθ.

[JEE Main 2019 · 12 Jan · Shift II]

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