Question

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

Answer 1

It is given that the curve x 2 =4y passes through the point ( 1,2 ).

The equation of the given curve is,

y 2 =4x(1)

Differentiate equation (1).

2y dy dx =4 dy dx = 2 y

Write the equation for the slope of the tangent at point ( 1,2 ) on the curve.

dy dx = 2 2 =1

Since the slope of the normal is negative of reciprocal of the slope of tangent, therefore write the expression for the slope of the normal at the point ( 1,2 ) on the curve.

−1 m = −1 1 =−1

The formula for the equation of normal is,

y− y 1 =m( x− x 1 )

Substitute 2 for y 1 and 1 for x 1 in the above equation.

y−2=−1( x−1 ) y−2=−x+1 x+y=3 x+y−3=0

Thus, the equation of normal to the given curve is x+y−3=0.