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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2019 · 12 January · Evening Shift · Mathematics · Parabola

The equation of a tangent to the parabola, x ^ 2 = 8y , which makes...

The equation of a tangent to the parabola, x2=8y{x}^{2} = 8y, which makes an angle θ\theta with the positive direction of x−x - axis, is

Options

  1. Ay=xtanI^¸+2cotI^¸y = xtanθ + 2cotθ
  2. By=xtanI^¸−2cotI^¸y = xtanθ - 2cotθ
  3. Cx=ycotI^¸+2tanI^¸x = ycotθ + 2tanθ
  4. Dx=ycotI^¸−2tanI^¸x = ycotθ - 2tanθ

Correct answer

C. x=ycotI^¸+2tanI^¸x = ycotθ + 2tanθ

Step-by-step solution

As we know, if the line y=mx+cy = mx + c is a tangent to x2=8y=4×2×y{x}^{2} = 8y = 4 \times 2 \times y, then c=−2m2c = - 2{m}^{2}.

So, equation of tangent with slope mm is

y=mx−2m2...(i)y = mx - 2{m}^{2} . . . \left(i\right)

Since, tangent is inclined at θ\theta with the positive x−x - axis, i.e. m=tanθm = tan \theta.

Equation of tangent is y=tanθ.x−2tan⁡2θy = tan\theta . x - 2\tan^{2} \theta

⇒ycotθ=x−2tanθ⇒ ycot \theta = x - 2tan\theta

⇒x=ycotθ+2tanθ⇒ x = ycot\theta + 2tan\theta

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