Angle between the curves without differentiation $y^2=4(x+1)$ and $x^2=4(y+1)$
Are the two curves, the answer online uses calculus and we have not been taught that.
can this be solved any other way?
Thanks in advance.

Write the product of the equations of the parabolas around the intersection points $(2\pm 2\sqrt 2, 2\pm 2\sqrt2):$ $-16[(x-(2\pm 2\sqrt{2}))^2\pm \sqrt{2}(x-(2\pm 2\sqrt{2}))(y-(2\pm 2\sqrt{2}))+(y-(2\pm 2\sqrt{2}))^2]+$ higher order terms, and use the $\tan \theta = {2 \sqrt{h^2-ab} \over a+b},ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0:$ $$\tan \theta={2 \sqrt{(\sqrt{\pm 2})^2-1} \over 1+1}=1.$$
