Find a curve whose distance of every tangent from the origin $ON$ is equal to the $x$ axis coordinate of the point of intersection between the curve and that tangent $OU$.
How to set up the graph for this kind of problems in general?
How to form an ODE?
How to choose constant of integration and the resulting graph?
EDIT:
Here is the sketch:
Line $OB$ is orthogonal to tangent. The condition is that $OA=OB$
On
I understood the problem simply this way as commented also earlier( the curve and tangent are same):
$$ p=r \sin \psi ,\, x= r \cos \theta ,\, \sin \psi= \cos \theta \rightarrow \theta + \psi = \pi/2 = \phi$$
which are straight lines $ x= const =p $ all parallel to $y-$axis
$$ \phi=\pi/2,3\pi/2, $$
the latter applies for straight lines to the left of the y-axis.They are directly verified above. Polar equation of curve ( = straight line) is $ r= p \sec \theta $
As a another ODE setup exercise I suggest following slightly more involved problem for you:
Find the curve whose pedal length $p$ ( length of perpendicular dropped from origin onto tangent equals $y$ coordinate length at point of tangency.
Solution
$$ r \sin \psi = y= r \sin \theta $$
$$ \rightarrow \psi = \theta,\, \tan \psi = \tan \theta $$
Draw differential triangle
$$ r \,d \theta = dr \, \tan \theta $$
Integrate
$$ \log \sin \theta = \log r + \log c $$
Simplify to polar form equation getting all circles equation through origin :
$$ r = 2 R \cos (\theta - \alpha ) $$
where $ R, \alpha $ are arbitrary constants of rotated circles around $z$ axis with arbitrary size and orientation ( its radius and rotation of diameter from position of center on $x$ axis.)
EDIT1
It can be in fact be generalized using same method.Choose a kite shaped cyclic quadrilateral, the red line is the required curve and its own tangent line for an arbitrary inclination $ 2 \theta $ between pedal normal and x-axis drawn inclined conveniently. The pedal length $ r \sin \psi =x = p$ as shown.
On
This is not a solution, but I thought having a picture could help fixing ideas.
For every point $A,B,C,...$ on x-axis, draw a circle centerred in $O$.
Since the distance to the tangent is calculated by the perpendicular to that tangent, then it has also to be tangent to the circle. For instance $[O,g]\perp[g,GG]$ is represented on the picture in pink color.
The green curve is the one we try to compute, while the orange curve on the left is made from points where the distance to the tangent is calculated.
I started from an arbitrary point $a$ and adjusted then $b,c,...,h$ such that there is tangency with $[AA,BB]$ and so on, and that gave the green curve.
Given the answer by Yves Daoust, I added the circle in brown, and effectively it fits my construction. Well done !
On
The equation of the tangent at abscissa $x$ is
$$Y-y(x)-y'(x)(X-x)=0$$ so that the distance to the origin is
$$\frac{|0-y-y'\cdot(0-x)|}{\sqrt{1+y'^2}}$$ and is known to equal $x$.
Squaring and rearranging,
$$y^2-2xyy'+x^2y'^2=x^2+x^2y'^2,$$
$$y^2-2xyy'=x^2,$$
$$\left(\frac{y^2}x\right)'=-1.$$
Then after integration
$$y^2=x(C-x).$$
Discussion:
As the distance is an absolute value, both signs in $y=\pm\sqrt{x(C-x)}$ are valid and we can leave the equation in its quadratic form.
EDIT: Let me rephrase the question, if I have understood it right. Write the ODE that corresponds to the family of curves that fulfill this condition: if we consider the tangent to one such curve at a point $x_0>0$, then the (orthogonal) distance from this tangent line to the origin is precisely $x_0$.
Consider a curve $y=f(x)$ and a point $x_0$. Then, the tangent at $(x_0,f(x_0))$ is $y=f(x_0)+f'(x_0)(x-x_0)$. We know that the distance from the origin to this line (see e.g. https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Line_defined_by_an_equation) is: $$ distance=\frac{|f(x_0)-f'(x_0)x_0|}{\sqrt{1+(f'(x_0))^2}} $$ But the abscissa of the point of intersection of the tangent to the curve is $x_0$ itself, so the requested ODE is, dropping the subindex of $x_0$: $$ x=\frac{|f(x)-f'(x)x|}{\sqrt{1+(f'(x))^2}} $$ or substituting $y=f(x)$ for clarity: $$ x=\frac{|y-y'\,x|}{\sqrt{1+(y')^2}} $$ Squaring and simplifying yields: $$ x^2-y^2+2y y' x=0 $$ or $$ y'=\frac{y}{2x} - \frac{x}{2y} $$ which is a well known Bernouilli's equation with solution $C x=x^2+y^2$ (see https://www.wolframalpha.com/input/?i=y%27%3Dy%2F(2x)-x%2F(2y))