Question

Find the equation of the normal at (1, 2) to the curve x² = 4y.

Answer 1

Given:

(x1, y1) = (1, 2)

Equation of the curve with respect to x.

`d/dx(x^2) = d/dx(4y)`

2x = `4(dy/dx)`

`dy/dx = (2x)/4 = x/2`

The slope of the tangent at the point (1, 2) is: 

m_T = `(dy/dx)_"(1,2)" = 1/2`

`m_N = − 1/(m_T)`

`m_N = − (1)/(1/2) = −2`

The equation of a line passing through (x₁, y₁) with slope is given by:

y − y₁ = m(x − x₁)

y − 2 = −2(x − 1)

y − 2 = −2x + 2

Now, rearrange the equation into the general form (Ax + By + C = 0):

2x + y − 2 − 2 = 0

2x + y − 4 = 0