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

1 Answer
Jun 11, 2018

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