Question

Find the equation of the normal to the curve x² = 4 y which passes through the point (1, 2).

Solution

The equation of the curve is x² = 4y ...(1)
$therefore space space space space space space space space space space straight y space equals space straight x squared over 4 space space space space space space space rightwards double arrow space space space space space dy over dx space equals space fraction numerator 2 straight x over denominator 4 end fraction space equals space straight x over 2$
Let normal at (h, k) pass through (1, 2).
Since (h, k) lies on (1)
$therefore space space space space straight h squared space equals space 4 straight k$ ...(2)
Slope of tangent at (h, k) = $straight h over 2$
$therefore space space space space space space slope space of space normal space at space left parenthesis straight h comma space straight k right parenthesis space equals space minus 2 over straight h$
Equation of normal at (h, k) is $straight y minus straight k space equals space minus 2 over straight h left parenthesis straight x minus straight h right parenthesis$
$because space space space space space space space it space passes space through space left parenthesis 1 comma space 2 right parenthesis space space space space space space space space space space space space space space space space space space space space space because space space space space space space space 2 minus straight k space equals space minus 2 over straight h left parenthesis 1 minus straight h right parenthesis$
$or space space space space space 2 straight h minus hk space equals space minus 2 plus 2 straight h space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight h space straight k space equals space 2 From space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space get comma space straight h open parentheses straight h squared over 4 close parentheses space equals space 2. therefore space space space space space space space space straight h squared space equals space 8 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight h space equals space 2 space space space space space space space space space space space therefore space space space straight k space equals space 2 over straight h space equals space 2 over 2 space equals space 1 therefore space space space space space space equation space of space normal space is space straight y minus 1 space equals space minus 2 over 2 left parenthesis straight x minus 2 right parenthesis or space space space space space space straight y minus 1 space equals space minus straight x plus 2 space space space space space space space space space space space space space space or space space space space space space space straight x plus straight y minus 3 space equals space 0$

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Continuity And Differentiability

Find the equation of the normal to the curve x² = 4 y which passes through the point (1, 2).

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Continuity And Differentiability

Find the equation of the normal to the curve x2 = 4 y which passes through the point (1, 2).

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the equation of the normal to the curve x² = 4 y which passes through the point (1, 2).