Conic Sections · JEE Preparation Mathematics
The equation of the tangent to the parabola y^2 = 16x at the point (4, 8) is:
Practice question from Conic Sections (JEE Preparation Mathematics), with the correct answer and a step-by-step solution.
hard · MCQ · Q-63EA0F
The equation of the tangent to the parabola y^2 = 16x at the point (4, 8) is:
- A.y = x + 4Correct answer
- B.y = 2x + 4
- C.y = x - 4
- D.y = 4x + 8
Solution
For , , so . The tangent at is , giving or .
More from Conic Sections
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