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📐 Mathematics  ·  Limits and Derivatives  ·  JEE

Find the equation of normal to curve xy = 4 at point (2,2).

Answer: y = x.

  • A y = x
  • B y = -x + 4
  • C y = x + 4
  • D y = -x

Correct answer: A. y = x

Explanation: Implicit differentiation of xy = 4: y + xy' = 0, so y' = −y/x = −1 at (2,2). Tangent slope = −1, so normal slope = 1. Normal line: y − 2 = 1·(x − 2), giving y = x.

xyP (a, f(a))Q (b, f(b))secant PQtangent at PAs Q slides toward P, secant approaches tangent

As point Q slides along the curve toward P, the secant line PQ rotates into the tangent line at P, whose slope is the derivative.

Concept context

Rate of change, differentiation rules, and applications

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