Jameson is correct but it's a bit more complicated than he implies.
Suppose a line is tangent to y= x2+ 1 at (x0,x02+ 1) and tangent to y= -x2 at (x1,-x12).
Any (non-vertical) line can be written as y= mx+ b. m is equal to the derivative of the functions at the given points: m= 2x0= -2x1 so x1= -x0. We must have x02+1= (2x0)x0+ 1 or b= 1- x02. We must also have -x12= (-2x1)x1+ b or b= x12. That is, b= x12= 1- x02. But since x1= -x0, x12= x02 so 1- x02= x02.
Solve that for x0 and then you can find m and b.
Because of the squares, there are, of course, two symmetric solutions,.