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

Explanation

Equation of given curve is : `y^(2) = 4x`
` rArr 2y(dy)/(dx) = 4 rArr (dy)/(dx) = 4/(2y) = 2/y`
`:." slope of tangent at point "(1, 2) is ((dy)/(dx))_(1, 2) = 2/2 = 1`
Slope of normal at point `(1, 2), = - 1/1 =- 1`
` :. ` equation of normal at point (1, 2),
` y-2 =- 1(x-1)`
` rArr y-2=-x+1 rArr x+y-3=0`