Find area bounded by parabola $y^2=2px,p\in\mathbb R$ and normal to parabola that closes an angle $\alpha=\frac{3\pi}{4}$ with the positive $Ox$ axis.

Question

Let $p=-2<0\Rightarrow y=\sqrt{-4x} \lor y=-\sqrt{-4x}\Rightarrow x\le 0 $.

Let $p=2>0\Rightarrow y=\sqrt{4x} \lor y=-\sqrt{4x}\Rightarrow x\ge 0 $.

For $p>0$ we can find the equation for normal to parabola such that it closes an angle $\alpha=\frac{3\pi}{4}$ with the positive $Ox$ axis, but for $p<0$ we can't find the normal.

Question: Is it correct that for $p<0$ we can't find normal to the given parabola?

How to find the equation of normal if the angle is given?

EDIT:

I am getting the negative result for total area. Here is how I integrated:

$A=A_1-A_2$ where $A_1$ is the area below $y^2=2px,p>0$ on interval $[-3p,p]$ integrating on $y$ axis. $A_2$ is the area below $y=-x+\frac{3p}{2},p>0$ on interval $[-3p,p]$integrating on $y$ axis.

$A=\int\limits_{-3p}^p \left(\frac{y^2}{2}-\frac{3p-2y}{2}\right)\mathrm dy$

$A_1=\int\limits_{-3p}^{p} \frac{y^2}{2}\mathrm dy=\frac{14p^3}{3},p>0$

$A_2=\int\limits_{-3p}^{p}\frac{3p-2y}{2} \mathrm dy=10p^2$

$A=A_1-A_2=\frac{2p^2(7p-15)}{2}$

Why is this wrong? It should be positive for every $p>0$, right?

Answer 1

Hint:

The situation for $p>0$ or $p<0$ is symmetric with respect to the origin. ( see the figure)

The point $A$ is the point where the tangent to the parabola forms an angle of $\pi/4$ with the $x-$axis, so you can find it solving $y'(x)=1$ for $y=\sqrt{2px}$ ( and find $A=(\frac{p}{2},p)$).

The line $AB$ has slope $m=-1$ ad contains $A$

can you do from this?

Finally: note that if you want the area of the region between the line and the parabola it is simpler to change axis with a transformation: $(x,y) \rightarrow (y,x)$.

Answer 2

Emilio Novati has already provided a good hint. This answer is to the following question in the EDIT :

Why is this wrong?

Since $y^2=2px\Rightarrow x=\frac{y^2}{2p}$, you should have $$A_1=\int_{-3p}^{p}\frac{y^2}{2\color{red}{p}}\mathrm dy=\frac{14}{3}p^{\color{red}{2}}$$

Then, note also that $$A=A_{\color{red}{2}}-A_{\color{red}{1}}=10p^2-\frac{14}{3}p^2=\frac{16}{3}p^2$$