Through the origin, a line is drawn parallel

Through the origin, a line is drawn parallel to the tangent at a point $T$ of a curve. Find the locus of the intersection of this line with the vertical through $T$, for 1° $y^2=2px$ and 2° $x^2+y^2=a^2$.

Step 1: Part 1°: the parabola $y^2=2px$

With $T=(x_0,y_0)$, the slope is $y'=p/y_0$. The line through the origin parallel to the tangent is $y=\dfrac{p}{y_0}x$; the vertical through $T$ is $x=x_0=y_0^2/(2p)$. Their intersection:

\[ x_M = \frac{y_0^2}{2p}, \qquad y_M = \frac{p}{y_0}\cdot x_M = \frac{y_0}{2}. \tag{1} \]

Eliminating $y_0=2y_M$: $x_M=(2y_M)^2/(2p)=2y_M^2/p$. Writing $(x,y)$ for the running point: \[y^{2} = \frac{p}{2}\,x. \tag{2}\]

A parabola with the same vertex and axis, but half the original semi-latus rectum.

Step 2: Part 2°: the circle $x^2+y^2=a^2$

Here $y'=-x_0/y_0$; the line through the origin parallel to the tangent is $y=-\dfrac{x_0}{y_0}x$, and the vertical through $T$ is $x=x_0$. Their intersection: \[x_M = x_0, \qquad y_M = -\frac{x_0^2}{y_0}. \tag{3}\]

So $x_0=x_M$, $y_0=-x_M^2/y_M$; substituting into $x_0^2+y_0^2=a^2$ and clearing denominators: \[x_M^2\!\left(1+\frac{x_M^2}{y_M^2}\right)=a^2 \;\Longrightarrow\; x_M^2(x_M^2+y_M^2) = a^2y_M^2. \tag{4}\]

Writing $(x,y)$ for the running point: \[x^{2}(x^2+y^2) = a^2y^2. \tag{5}\]

Step 3: The polar form

Setting $x=\rho\cos\varphi,\ y=\rho\sin\varphi$ in (5): $\rho^4\cos^2\varphi=a^2\rho^2\sin^2\varphi$, so, dividing by $\rho^2\cos^2\varphi$, \[\rho^{2} = a^2\tan^2\varphi \;\Longrightarrow\; \rho = a\,|\tan\varphi|. \tag{6}\]

Step 4: Verification

\[p=3,\ y_0=2.5\text{: }x_M=y_0^2/(2p)\approx1.04167,\ y_M=y_0/2=1.25,\ y_M^2=\tfrac p2x_M\approx1.5625\ \checkmark\]

\[a=2,\ \varphi=0.7\text{: }\rho=a\tan\varphi\approx1.68300\text{, and }x^2(x^2+y^2)-a^2y^2=0\text{ to machine precision}\ \checkmark\]