Using any parabola, find $$ such that it is tangent to the line $=$.

Question

Using any parabola, find $c$ such that it is tangent to the line $y=x$.

I am not sure what this question means. I thought tangent lines were straight? How can a parabola like $x^2 + x + 1$ be "tangent" to anything?

What is this problem asking me to do, and how can I do it?

This is from a question my professor admitted was a typo. It originally said, "Find $c$ such that the parabola is tangent to the line $y=x$". Since that was unsolveable, I emailed him and asked him to clarify, and he said, "feel free to work the problem with the parabola $x^2+x+1$, though any parabola you choose would do fine.."

I then asked on this website how to find $c$ such that $f(x)=x^2+x+1$ is tangent to the line $y=x$. Nobody seemed to know what this meant either, I got downvoted and deleted the question, so I'm asking a different question now.

I have no idea what this question is even supposed to look like, so please be understanding here.

Answer 1

A parabola $y=ax^2+bx+c$ will be tangent to a line $y=mx+q$ iff the two have exactly one point in common. This amounts to saying that the equation $$ax^2+bx+c=mx+q\implies ax^2+(b-m)x+(c-q)=0$$ has exactly one solution, i.e. in terms of the discriminant, that $$(b-m)^2-4a(c-q)=0$$ In your case the line is just $y=x$ which makes the general formula simpler $$(b-1)^2-4ac=0$$ hence you see that you need to find any triplet $a,b,c$ such that $$(b-1)^2=4ac.$$ Of course there are infinitely many such triplets.

---

If, instead, you want a parabola that is tangent to a line $y=mx+q$ at certain point $(x_0,y_0)$ then indeed you can use the derivative which results in the system $$\begin{cases}ax_0^2+bx_0+c=y_0\\ 2ax_0+b=m\\ \end{cases}$$ and adding a last condition will tell you the one parabola which you are looking for.

Answer 2

For the parabola $y = x^2$ the slope is $1$ where the derivative $2x = 1$. Hence at $x = 0.5, y = 0.25$ the slope is $1$. Hence if we raise the parabola by $0.25$ by changing it to $y = x^2 + 0.25$, we have a tangent of the line and the parabola $y = x^2 + 0.25$ at $(0.5, 0.5)$.