I am trying to find the curve passing through the point $(1,2)$ given that the the length of the perpendicular drawn from the origin to the normal at any point is equal numerically to the ordinate (fancy y coordinate) of the point.
This condition seems to force the curve to be something like a circle, but I am not sure. I am also not sure how to put the condition into math: At first glance this would seem to imply that $$ \sqrt{x^2+y^2}=y\Rightarrow x^2+y^2=y^2\Rightarrow x=0 $$ Which seems not true.
I could also use some help unpacking the perpendicular to the normal at any point part, this doesn't seem to add anything to the problem since it seems to be saying the shortest distance to the point. Is that correct?
The distance from the origin to the line $Ax+By+C=0$ is given by $$ d = \frac{|C|}{\sqrt{A^2+B^2}}. $$ To see this, note that if neither $A$ nor $B$ is zero, then the area of the triangle formed by the origin and the $x$ and $y$ intercepts is given by $\frac{1}{2}\frac{|C|}{|A|}\frac{|C|}{|B|}$ by taking the bases on the x-axis, while it is also given by $\frac{1}{2}d\sqrt{\frac{C^2}{A^2}+\frac{C^2}{B^2}}$ by considering the base as the hypotenuse instead. The formula follows by equating these quantities.
If $(x_0,y_0)$ is a point on a curve and $m$ is the slope at that point, then the tangent line to the curve at $(x_0,y_0)$ is given by $$ y- y_0 = m(x-x_0)\implies -mx + y + (mx_0-y_0) = 0. $$ The distance from the origin to that tangent line is thus $$ d = \frac{|mx_0-y_0|}{\sqrt{m^2+1}} $$ and the requirement given in the problem is that $$ y_0 = d = \frac{|mx_0-y_0|}{\sqrt{m^2+1}}. $$ This implies that \begin{align} y_0\sqrt{m^2+1} = |mx_0-y_0| \implies y_0^2(m^2+1) &= (mx_0-y_0)^2 \\ \implies y_0^2m^2 + y_0^2 &= m^2x_0^2 - 2my_0x_0 + y_0^2 \\ \implies 0 &= m((x_0^2-y_0^2)m - 2y_0x_0). \end{align} We thus have $$ m = 0\text{ or } m = \frac{2y_0x_0}{x_0^2-y_0^2}. $$ Now $m = 0 $ implies the constant curve $y=2$, which obviously works. The alternative condition yields the differential equation $$y' = \frac{2yx}{x^2-y^2} = \frac{2}{\frac{x}{y} - \frac{y}{x}}. $$ Letting $z = \frac{y}{x}$, we have $$ z' = \frac{y'}{x} - \frac{y}{x^2} = \frac{1}{x}\left(y' - \frac{y}{x}\right) = \frac{1}{x}\left(\frac{2}{\frac{1}{z}-z} - z\right) = \frac{z}{x}\left(\frac{2}{1-z^2}-1\right) = \frac{z}{x}\left(\frac{1+z^2}{1-z^2}\right). $$ Separating variables, we get $$\frac{1-z^2}{z(1+z^2)}z' = \frac{1}{x}. $$ Now $$\frac{1-z^2}{z(1+z^2)} = \frac{1+z^2}{z(1+z^2)} - \frac{2z^2}{z(1+z^2)} = \frac{1}{z} - \frac{2z}{1+z^2}$$ and hence $$\frac{1}{x} = \frac{1-z^2}{z(1+z^2)}z' = \left(\frac{1}{z} - \frac{2z}{1+z^2}\right)z' = \left(\log z - \log(1+z^2)\right)'. $$ This implies that $$\log\left(\frac{z}{1+z^2}\right) = C + \log x \implies \frac{z}{1+z^2} = Cx$$ for some constant $C$. For $(x_0,y_0) = (1,2)$, we have $z_0 = 2$, yielding $$\frac{2}{1+2^2} = C(1)\implies C = \frac{2}{5}. $$ At this point, we could solve for $z$ and thus solve for $y$. However, we can instead introduce polar coordinates to simplify the expressions above. Letting $x = r\cos\theta$, $y=r\sin\theta$, we then get $z = \tan\theta$, and hence \begin{align} \frac{\tan\theta}{1+\tan^2\theta} &= \frac{2}{5}r\cos\theta \\ \implies \frac{\tan\theta}{\sec^2\theta} &= \frac{2}{5} r\cos\theta \\ \implies \tan\theta\cos^2\theta &= \frac{2}{5}r\cos\theta \\ \implies \tan\theta\cos\theta &= \frac{2}{5}r \implies r = \frac{5}{2}\sin\theta. \end{align} Thus, the desired curve is given by $r = \frac{5}{2}\sin\theta$. We can turn this into an equation in Cartesian coordinates by multiplying both sides by $r$, yielding $$ r^2 = \frac{5}{2} r\sin\theta\implies x^2+y^2 = \frac{5}{2}y. $$