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.
 A: 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".

Note: If this is not enough, i will come back with pictures and computations.
A: 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 $135°$.

A: Motivation:

*

*Changing solution sets:
Suppose we have a quadratic:
$$ P(x) = ( x-a)(x-b)$$
Suppose I wanted another function for which the roots are square of the polynomial before, then I plug $ g(x) = \sqrt{x}$ into P:
$$ P(g) = ( \sqrt{x} -a ) ( \sqrt{x} - b) \tag{1}$$
So plugging in the inverse function of the desired result we want.
2. Rotations:
Using complex numbers we can find to rotate a number by clockwise ninty degrees we simply need to multiply by $-i$ i.e: if $ z= x+iy$ then rotated clockwise by ninty degree about origin is $-iz = -ix+y$. Hence under rotation of ninty about origin, we have the following map:
$$(x,y) \to (y,-x) \tag{2}$$
3. The actual answer:
Consider your parabola:
$y^2=4(x+1)$
If I wanted to rotate the whole curve by ninty counter clock, I plug in the coordinates with inverse action applied(2) (i.e: ninty clockwise) , this leads to:
$$(-x)^2 = 4(y+1)$$
This gives:
$$x^2 = 4(y+1)$$
Now, if our two curves have an intersection, this this point is uneffected by this rotation and hence it must be that the tangent of the original curve must be rotated by ninty degree. Hence angle between intersection is ninty.
The key idea here is knowing how solution sets of the equation change when you plug in things and rotation and work back your way to the observation about perp tangents
A: Hint.
Calling
$$
\cases{
f(x,y) = y^2-4(x+1)\\
g(x,y) = x^2-4(y+1)
}
$$
we have $f(x,y) = g(y,x)$ then $f(x,y)\cap g(x,y)\in \{x=y\}$ and thus we can determine an intersection point $(x_0,y_0)$. Now the tangent line to $f(x_0,y_0)$ can be determined as follows: calling $y = y_0+m(x-x_0)$ we have that solving for $x$
$$
f(x,y_0+m(x-x_0)) = 0
$$
we will determine the intersections of $y=y_0+m(x-x_0)$ with $f(x,y)=0$. This gives
$$
x^* = \frac{m^2 x_0-m y_0+2\pm 2 \sqrt{m^2 x_0+m^2-m y_0+1}}{m^2}
$$
now as $y=y_0+m(x-x_0)$ is tangent, the intersection should be unique thus
$$
m^2 x_0+m^2-m y_0+1 = 0
$$
and solving for $m$ we get
$$
m_f = \frac{y_0\pm\sqrt{y_0^2-4(x_0+1)}}{2 (x_0+1)} = \frac{y_0}{2(x_0+1)}
$$
because $y_0^2-4(x_0+1)=0$.
Now with the same procedure for $g(x,y_0+m(x-x_0))=0$ we get
$$
m_g = \frac{x_0}{2}
$$
Now having the tangents $(m_f,m_g)$ we can calculate the angles...
NOTE
$m$ should be non null so in case of vertical solutions, the approach should be slightly different.
$$
\tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} = \frac{m_f-m_g}{1+m_f m_g}
$$
then
$$
\theta_f-\theta_g = \arctan\left(2\frac{y_0-x_0(x_0+1)}{x_0(y_0+4)+4}\right)
$$
A: 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 $\tan \theta = {2 \sqrt{h^2-ab} \over a+b},ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0:$ $$\tan \theta={2 \sqrt{(\sqrt{\pm 2})^2-1} \over 1+1}=1.$$
