Q: Analytic geometry and definite integrals problem... So, here's the problem:
We have a parabola $y^2=2px$ and a line which is perpendicular to parabola and forms the angle $\frac{3\pi}{4}$ with x axis. I have to find the area between the parabola and the line. I can't seem to find the point where that line intersects with y axis. Since $\tan(\frac{3\pi}{4})=-1$, which is the first derivative of the line equation (it's coefficient), that means the line equation looks like this: $y=-x+n$ How can I find n?

A: You need to find the point on the parabola where the normal has gradient $-1$, i.e. where the tangent has gradient 1. This is the point through which the line you have found must pass, and this will determine $n$.

A: First, find the equation of the general normal to the curve. It passes through $(x_0,y_0)=(y^2_0/(2p),y_0)$. Differentiating implicitly, we find
$$ 2y \frac{dy}{dx} = 2p, $$
so the gradient is $p/y_0$.
Since, as you note, the gradient of the normal has to be $-1$, the gradient of the parabola at the point of intersection should be $-(1/(-1)) = +1$. Hence $y_0=p$, and the equation of the line is therefore
$$ y-p = -(x-p/2), $$
or $x = 3p/2-y$. Now you need to find the other point of intersection of this line with the parabola, and then find the integral, which is going to be easiest if you integrate $x$ with respect to $y$.
