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Application Of Derivatives

Question
CBSEENMA12035075

Find the equation of the normal to the curve y2 = 4x at the point (1, 2).

Solution

The equation of curve is
y2 = 4x
Differentiating both sides w.r.t. x, we get,
                           2 straight y dy over dx space equals space 4 space space space space space space space or space space space dy over dx space equals space 2 over straight y
therefore space space space space at space left parenthesis straight x comma space straight y right parenthesis. space slope space of space tangent space space equals space 2 over straight y
therefore space space space space space at space left parenthesis 1 comma space 2 right parenthesis comma space slope space of space tangent space equals space 2 over 2 space equals space 1
therefore space space space space space slope space of space normal space space equals space minus 1 over 1 space equals space minus 1
therefore space space space space space equation space of space normal space at space left parenthesis 1 comma space 2 right parenthesis space is space
space space space space space space space space space space space space space space space space space space space space space space space space straight y space minus space 2 space equals space minus space 1 left parenthesis straight x minus 1 right parenthesis space space space space space space or space space space straight y minus 2 space equals space space minus straight x plus 1
or space space space space space space space space space straight x plus straight y minus 3 space equals space 0.