Find a curve that passes through point $A(2,0)$, such that a triangle defined with a tangent at an arbitrary point $M$, $0y$ axis, and with a line $OM$ is isosceles, where $OM$ is the triangle base.
Is this the right sketch?
EDIT:
We have that $OB=BM$.
$$OM: y=\frac{y_0}{x_0}x$$ $$MB: y=\frac{y_B-y_0}{x_B-x_0}x+\frac{y_0x_B-y_Bx_0}{x_B-x_0}$$ $$y=f'(x_0)x+(y_0-f'(x_0)x_0)$$ $$f'(x_0)=\frac{y_B-y_0}{x_B-x_0},y_0-f'(x_0)x_0=\frac{y_0x_B-y_Bx_0}{x_B-x_0}$$ $$OB: x=0$$
How to find coordinates of $M$ and $B$? How to form differential equation from $$f'(x_0)=\frac{y_B-y_0}{x_B-x_0}?$$
There is an evident solution: the circle with unit radius centered in $C(1,0)$ (see figure) because the 2 tangents issued from a point to a circle have equal lengthes.
Remark: We have to accept that infinite size triangles can be considered, in order that point $A$, in which there is a vertical tangent (that intersects $y$ axis at infinity), is included (see the remark of Emilio Novati).
An analytical approach is possible. Here is how.
Let $y=f(x)$ be the looked for equation.
Let $M(x_0,f(x_0))$ be the current point on the looked for curve.
Using the classical equation of the tangent at the curve in $M$, i.e.,
$$\tag{1}y=f(x_0)+f'(x_0)(x-x_0)$$
Setting $x=0$ in (1), we get the coordinates $(x_B=0,y_B=f(x_0)-f'(x_0)x_0)$ of point $B$.
Let us now write that $BO^2=BM^2$:
$$x_B^2+y_B^2=(x_0-x_B)^2+(f(x_0)-y_B)^2$$
$$(f(x_0)-f'(x_0)x_0)^2=x_0^2+f(x_0)^2$$
Expanding and reducing:
$$f(x_0)^2-2 f(x_0)f'(x_0)x_0-x_0^2=0$$
Whence:
$$\tag{2}f'(x_0)=\dfrac{f(x_0)}{2x_0}-\dfrac{x_0}{2f(x_0)}$$
a differential equation for which we can check that:
$$f(x_0)=\pm\sqrt{x_0^2-2x_0}$$
(the equation of the upsaid circle) is a solution.
But I confess that I am not able to prove rigorously that it is the unique solution of (2) such that $f(2)=0.$