A curve $C$ in the $x-y$ plane is such that the line joining the origin to any point $P(x,y)$ on the curve and the line parallel to the $y$ axis through P are equally inclined to the tangent to the curve at P. Find the differential equation of the curve $C$.
Slope of the line from origin to $P(x,y)$ will be $\frac{y}x = m1$
Slope of the line from $P$ parallel to y axis = $\tan(90) = m2$
I am having trouble proceeding from here.
On
Suppose the curve is given in parametric form $$ C: \vec{r} = \langle x(t), y(t) \rangle$$
At any time $t_0$, a point on the curve is given by
$$ P(x_0,y_0) = \langle x(t_0), y(t_0) \rangle $$
The vector tangent to $C$ at $P$ is $$ \vec{v} = \langle \dot x(t_0), \dot y(t_0) \rangle $$
The direction vector of the line $OP$ is given by
$$ \vec{u_1} = \langle x(t_0), y(t_0)\rangle $$
and the direction vector of the line $x=x_0$ (parallel to the $y$-axis) is simply the $y$ unit vector $$\vec{u_2} = \langle 0, 1 \rangle $$
These two vectors make the same angle with the tangent vector, so by the dot product rule we have
$$ \cos\theta = \frac{\vec{u_1} \cdot \vec{v}}{\vert\vec{u_1}\vert\vert\vec{v}\vert} = \frac{\vec{u_2}\cdot\vec{v}}{\vert\vec{u_2}\vert\vert\vec{v}\vert} $$ $$ \implies \frac{x\dot x + y\dot y}{\sqrt{x^2+y^2}} = \dot y $$
The equivalent ODE where $y = y(x)$ can be found by rearranging the above to get
$$ \frac{dy}{dx} = \frac{\dot y}{\dot x} = \frac{x}{\sqrt{x^2+y^2}-y} $$
The solution of this turns out to be
$$ y = \frac{x^2-a^2}{2a} $$
According to the specifications we have
$$ \frac{dy}{dx} = \tan\left(\frac{\arctan(\frac yx)+\frac{\pi}{2}}{2}\right) $$
hence
$$ \frac{dy}{dx} = \frac{\sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{y}{x}\right)\right)+\cos \left(\frac{1}{2} \tan ^{-1}\left(\frac{y}{x}\right)\right)}{\cos \left(\frac{1}{2} \tan ^{-1}\left(\frac{y}{x}\right)\right)-\sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{y}{x}\right)\right)} = \frac{y}{x}+\sqrt{\frac{y^2}{x^2}+1} $$
The solution to this DE is
$$ y = -x \sinh\left(C_0-\log x\right) $$