Tangent parabolas

Question

Given that $y^2=x$ and $x^2=y$ are both parabolas, what is the value of $c$ if $y^2=x$ and $x^2+c=y$ are both tangent?

What I have tried:

$$\begin{cases} y^2=x\\ x^2+c= y \end{cases}$$ Taking the tangent line equation I get $$\begin{cases} y-y_1=\frac 1 {2y_1} (x-x_1) \\ y-y_2=2x_2(x-x_2) \end{cases}$$ Comparing the equations getting $8x_2c-8x_2^3=1$, finding the value of $c$ such that this equation has only one real solution, am I on the right track? What should I do?

Answer 1

WLOG any point on $y^2=x$ will be $P(t^2,t)$

So, the derivative at $P,$ $$\dfrac{dy}{dx}=\dfrac1{2t}$$

The derivative of $y=x^2+c$ at $P$ is $$2t^2$$

So, we need $2t^2=\dfrac1{2t}\implies t=?$

$$\implies c=y-x^2_{(\text{at }P)}=t-t^4=?$$

Answer 2

HINT

Essentially, the question is asking us to equate the derivatives of the $2$ functions.

If you need any more help, please don't hesitate to ask. I have a full solution ready to type up and post.