Find the equation of a line that is tangent to both parabolas simultaneously

Question

Consider the parabolas $y=x^2$ and $y=x^2-2x+2$. How to find the equation of a line that is tangent to both of them at the same time? Please, walk me through the most intuitive solution.

Answer 1

Hints:

• Let $(a, a^2)$ and $(b, b^2 - 2b + 2)$ be the two points of tangency on the common tangent line.
• Independently compute the equation of both tangent lines.

You should get: $$ \begin{cases} y - a^2 = 2a(x - a) \\ y - (b^2 - 2b + 2) = (2b - 2)(x - b) \\ \end{cases} \iff \begin{cases} y = (2a)x - a^2 \\ y = (2b - 2)x - b^2 + 2 \\ \end{cases} $$

• Compare coefficients to obtain a system of two equations and two unknowns.

You should get: $$ \begin{align*} 2a &= 2b - 2 \\ -a^2 &= -b^2 + 2 \\ \end{align*} $$

• Solve the system and plug the values back in to obtain the equation of the tangent line.

Answer 2

Let the tangent line be $y = mx + b$. We know that both $mx +b = x^2$ and $mx+b = x^2-2x+2$ have one single solution, which implies that the discriminants of those quadratic equations are zero: $$ m^2+4b =0, \quad (m+2)^2-4(2-b)=0 $$

This leads to $m=1$ and $b = -\frac 14$.