Tangent Line(s) common to two parabolas?

y = x² - 2x + 5

dy/dx = 2x - 2

Let P be a point on this first given curve: P(t, t² - 2t + 5)

The tangent line through P:

y - (t² - 2t + 5) = (2t - 2)(x - t)

y - t² + 2t - 5 = 2tx - 2t² - 2x + 2t

y = 2tx - 2x - t² + 5

Substitute that into the second given equation.

y = -x²/2 + 2x + 1

2tx - 2x - t² + 5 = -x²/2 + 2x + 1

x²/2 + 2tx - 4x - t² + 4 = 0

x² + (4t - 8)x + (8 - 2t²) = 0

The line is tangent to the second parabola, so the discriminant of this quadratic equation has to be zero.

(4t - 8)² - 4(1)(8 - 2t²) = 0

(2t - 4)² - (8 - 2t²) = 0

4t² - 16t + 16 - 8 + 2t² = 0

6t² - 16t + 8 = 0

3t² - 8t + 4 = 0

(3t - 2)(t - 2) = 0

t = 2/3 or t = 2

Substitute those into the tangent line equation.

y = 2tx - 2x - t² + 5

(2 - 2t)x + y + t² - 5 = 0

2x - y + 1 = 0

or

6x + 9y - 41 = 0