For example, parabola y = x 2+3.

The points outside the parabola are (1, 1).

Let the tangent point on the parabola be (a, a 2+3)

Derive parabola: y'(x)=2x

So:

Tangent slope k = y' (a) = 2a = (a 2+3-1)/(a-1)

So: a 2+2 = 2a 2-2a.

So: a 2-2a = 2.

So: (A- 1) 2 = 3.

Solution: a= 1+√3 or a= 1-√3.

So: k=2a=2+2√3 or 2-2√3.

So: the tangent is y- 1=k(x- 1), just substitute it.