Find the locus of the feet of the perpendiculars dropped

Find the locus of the feet of the perpendiculars dropped from the origin onto the tangents of the hyperbola $x^2-y^2=a^2$.

Step 1: Tangent at $T=(x_0,y_0)$

With $y'=x_0/y_0$, the tangent is $x_0x-x_0^2=y_0y-y_0^2$, i.e.

\[ x_0x - y_0y = x_0^2-y_0^2 = a^2. \tag{1} \]

Step 2: Perpendicular from the origin

Slope $-y_0/x_0$, through the origin: \[y_0x + x_0y = 0. \tag{2}\]

Step 3: The foot $M$

Substituting $y=-\frac{y_0}{x_0}x$ from (2) into (1), with $r^2:=x_0^2+y_0^2$: \[x_M = \frac{a^2x_0}{r^2}, \qquad y_M = -\frac{a^2y_0}{r^2}. \tag{3}\]

Step 4: Eliminate $x_0,y_0$

From (3): $x_0=r^2x_M/a^2$, $y_0=-r^2y_M/a^2$, and $x_M^2+y_M^2=a^4/r^2$ so $r^2=a^4/(x_M^2+y_M^2)$. Substituting into $x_0^2-y_0^2=a^2$: \[(x_M^2-y_M^2)\,r^4 = a^6 \;\Longrightarrow\; (x_M^2-y_M^2)\cdot\frac{a^8}{(x_M^2+y_M^2)^2} = a^6. \tag{4}\]

Dividing by $a^2$ and writing $(x,y)$ for $M$: \[(x^2+y^2)^2 = a^2(x^2-y^2). \tag{5}\]

In polar coordinates: $\rho^2=a^2\cos2\varphi$.

Step 5: Verification

\[a=3,\ x_0=5.5\ (y_0=\sqrt{x_0^2-a^2}\approx4.60977)\text{: foot }M\text{ satisfies }(x_M^2+y_M^2)^2-a^2(x_M^2-y_M^2)\approx0\ \checkmark\]

\[x_0=4.0,\,10.0\text{: same identity holds to machine precision in both cases}\ \checkmark\]