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Question 310012
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Application of Derivatives: JEE Advanced MCQ

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JEE AdvancedHardQuestion 310012

Question

The equation of the normal to y = x² − 1 at the point (1, 0) is:
  1. A
    x + 2y − 1 = 0
    Correct
  2. B
    x − 2y + 1 = 0
  3. C
    2x − y − 2 = 0
  4. D
    x = 1

Correct answer

x + 2y − 1 = 0

Explanation

y' = 2x → at (1,0): slope of tangent = 2. Slope of normal = −1/2. Normal: y − 0 = −(1/2)(x − 1) → 2y = −x + 1 → x + 2y − 1 = 0.

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