Possible values of slope of a line A variable line $y=mx+4$ either touches or intersects the curves $y=x^2$ and $x^2=1+\frac{y^2}8$ in total three distinct points, the question is to find out the possible values of $m$.
This question came in an examination in which calculators were not allowed. I tried to plot the graph manually but failed to find out the answer. I tried it on desmos and here is a snapshot of the graph:

Is there a way to find out value of $m$ theorecticaly by applying derivatives? I tried it several times but couldn't get. Any help shall be appreciated. Thanks. 
 A: If it touches at $3$ distinct points, then it must intersect at exactly one of the intersections of those two curves. So, we must find that intersection. That is $$y=x^2=1+\frac{y^2}{8}\implies \frac{1}{8}y^2-y+1=0\implies y=\dfrac{1\pm\sqrt{1-\frac{1}{2}}}{\frac{1}{4}}=4\pm2^{3/2}$$
For simplicity, take the positive of this value and the positive $x$ value when calculating $$y=x^2\implies x=\pm\sqrt{4\pm2^{3/2}}\stackrel{\text{pos.}}{\to}\sqrt{4+2^{3/2}}$$
Note that this corresponds to the upper right intersection point of the two curves. This is a point that the line must pass through. We only need one point to fully determine the equation of the line if its $y$-intercept must be $4$: $$y=mx+4\implies 4+2^{3/2}=m\sqrt{4+2^{3/2}}+4\implies m=\frac{2^{3/2}}{\sqrt{4+2^{3/2}}}$$
There are $3$ other lines that satisfy your conditions (and all have different slopes). One of them is just this line reflected over the $y$-axis (one of the other intersections). The other two just go through the other intersections.
