Question

How do you find the equation of the tangent and normal line to the curve `y=6-x^2` at (2,2)?

Answer 1

Tangent:

`y = -4x+10`

Normal:

`y = 1/4x + 3/2`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is `-1`

We have:

`y = 6-x^2`

First let us check that `(2,2)` lies on the curve:

`x=2 => y=6-4=2`

Then differentiating wrt `x` gives us:

`dy/dx = -2x`

When `x = 2 => dy/dx = -4`

So the tangent passes through `(2,2)` and has gradient `m_T=-4`, and the normal has gradient `m_N=1/4` so using the point/slope form `y-y_1=m(x-x_1)` the equation we seek are;

Tangent:

`y - 2 = -4(x-2)`
`:. y - 2 = -4x+8`
`:. y = -4x+10`

Normal:

`y - 2 = 1/4(x-2)`
`:. y - 2 = 1/4x-1/2`
`:. y = 1/4x + 3/2`

We can confirm this solution is correct graphically: