Angle between the curves without differentiation

The two curves intersect in $\Bbb R^2$ in the two points $P_\pm=(2\pm\sqrt 2, 2\pm\sqrt2)$. Both points are on the line $x=y$. So it is enough to get the angle of the tangent of each parabola in $P_\pm$ with the line $x=y$.

So let us consider first parabola $4(y+1)=x^2$. Its focus is the origin. Take a look for instance at:

https://en.wikipedia.org/wiki/Parabola

It is also easy to see this algebraically. The squared distance from a point $P=(x,y)$ to the focus $F=(0,0)$ is $x^2+y^2$, and the squared distance from $P$ to the horizontal line $(d)$ given by $y=-2$ (diretrix) is $(y+2)^2$. So the parabola, the geometric locus of all points equally far from $F$ and $(d)$ has the equation $x^2+y^2=(y+2)^2$. This is our parabola.

Using the "optical properties", a ray comming verically from "infinity" to the point $P_\pm$ is reflected (in the "wall of the parabola", i.e. in the tangent in that point to the "wall"), and the reflection passes through the focus. This means that the we know the angle of the tangent in $P_\pm$ at the parabola.

(Arguably, we still have to show this optical property without differential calculus... else this would be a "cheating answer", since we do not use differentiation, but an argument, that may use it. Is this an issue?)

For the other parabola, take a "horizontal rays".

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Note: If this is not enough, i will come back with pictures and computations.

As we can see the angle between the two curves at point $(2+2\sqrt 2,2+2\sqrt2)$ is twice the angle formed by the tangent and $x$-axis and the line $y=x$.

The tangent has gradient $f'(2+2\sqrt 2)= \sqrt{2}+1$

Thus the angle formed with $x$-axis is $\arctan(\sqrt 2+1)=67.5°$

Less the $45°$ of the angle bisector $y=x$ we see that the curves form an angle

$2(67.5°-45°)=45°$

In a similar way the angle formed by the two curves at the other intersection point is $135°$.