# 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?

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.

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

$$y = -x \sinh\left(C_0-\log x\right)$$

• An explanation would be nice – Dylan Jun 22 '18 at 6:35
• @Dylan Follows a graphic explanation. – Cesareo Jun 22 '18 at 9:03
• seems that you lost a factor $1/2$ at the beginning: shouldn't it be $\tan((\frac{\pi}{2}+\arctan(\frac yx))/2)$ ? – G Cab Jun 22 '18 at 9:18
• This doesn't look right. You want the angle between the vertical and the tangent, to be equal to the angle between the origin line and the tangent, i.e. both lines should make the same angle with the tangent. – Dylan Jun 22 '18 at 9:21
• According to my procedure the angle between the vertical and the line passing by the origin should be equal to the tangent angle with the horizontal line. – Cesareo Jun 22 '18 at 9:23