Question

What are the equations of the tangent lines out P(2,1) to the parabola with equation `y = x^2`? Thank you!

Answer 1

Two tangents

• `y = 2(2-sqrt 3)x-7+4sqrt3`
• `y = 2(2+sqrt 3)x-7-4sqrt3`

Explanation:

The tangent to a curve `y(x)` at the point `(x_0,y_0)` is given by

`y-y_0 = |dy/dx|_{x=x_0}(x-x_0)`

For `y=x^2` this becomes

`y-y_0 = 2x_0 (x-x_0)`

with `y_0 = x_0^2`, so that

`y = 2x_0 x-x_0^2`

For the tangent to go through `(2,1)` we must have

`1 = 4x_0-x-x_0^2 implies`
`x_0^2-4x_0+1 = 0 implies`
`(x_0-2)^2=3 implies`
`x_0 = 2 pm sqrt 3`

So, there are two tangents to the parabola from `(2,1)` to the parabola `y=x^2` and they are

• for `x_0 = 2-sqrt 3`, `y = 2(2-sqrt 3)x-7+4sqrt3`
• for `x_0 = 2+sqrt 3`, `y = 2(2+sqrt 3)x-7-4sqrt3`