Equation of a line that is tangent to 2 curves

Question

I'm given the question

Is there a straight line that is tangent to both the curves $y = x^2$ and $y = x^2 + 2x+2$? If so, find its equation. If not, why not?

I'm not entirely clear as to where to start with this form of question. Any guidance would be great!

Answer 1

Let $y=mx+c$ be the a tangent of $y=x^2$

Let us find the abscissa of intersection

$$x^2=mx+c\iff x^2-mx-c=0\ \ \ \ (1)$$

For tangency, $$(-m)^2-4(-c)=0\iff 4c=-m^2$$

Similarly for tangency of $$y=mx+c$$ with $$y=x^2+2x+2$$

$$(2-m)^2=4(2-c)\ \ \ \ (2)$$

Comparing the values of $c,$ $$m^2-4m+4=8+m^2$$

$m=\infty,-1$

Answer 2

Let $(t,t^2)$ and $(p,p^2+2p+2)$ be touching points.

Thus, $$2t=2p+2$$ and $$\frac{p^2+2p+2-t^2}{p-t}=2t,$$ which gives $t=-\frac{1}{2}$, $p=-\frac{3}{2}$ and the answer: $$y=-x-\frac{1}{4}.$$

Answer 3

Hint:

Tangency to a curve means that at some point you achieve the same ordinate and same slope.

Let the equation of the common tangent be $y=mx+p$, then

$$mx_0+p=x_0^2,\\m=2x_0,$$ and

$$mx_1+p=x_1^2+2x_1+2,\\m=2x_1+2.$$

Discuss the solutions of this system of four equations in four unknowns. (Don't be afraid, elimination of the unknowns is easy.)

Substract 4 from 2, giving $x_0-x_1=1$. Then subtract 3 from 1, giving $\color{green}m\cdot1=1\cdot(x_0+x_1)-2x_1-2=\color{green}{-1}$. Next $x_0=-1/2$ and $\color{green}p=x_0^2-mx_0=\color{green}{-1/4}$.