Show that all normals to the curve x=2at/(1+t²)

Show that all normals to the curve $x=\dfrac{2at}{1+t^{2}},\ y=\dfrac{a(1-t^{2})}{1+t^{2}}$, $a>0$, pass through a single point.

Step 1: Identify the curve

Computing $x^{2}+y^{2}$ directly (the numerator collapses via $4t^2+(1-t^2)^2=(1+t^2)^2$): \[x^{2}+y^{2} = \frac{4a^{2}t^{2}+a^{2}(1-t^{2})^{2}}{(1+t^{2})^{2}} = \frac{a^{2}(1+t^{2})^{2}}{(1+t^{2})^{2}} = a^{2}. \tag{1}\]

So the curve is nothing but the circle of radius $a$ centred at the origin, traced by $t=\tan\dfrac\theta2$ (the usual rational parametrisation).

Step 2: Parametric derivatives

\[ \dot x = \frac{2a(1-t^{2})}{(1+t^{2})^{2}}, \qquad \dot y = \frac{-4at}{(1+t^{2})^{2}}, \qquad \frac{dy}{dx} = \frac{-2t}{1-t^{2}}\ \ (t\neq\pm1). \tag{2} \]

Normal slope: $k_{n} = \dfrac{1-t^{2}}{2t}$.

Step 3: The normal at parameter $t$ passes through $O$

Writing the normal at $P=\left(\dfrac{2at}{1+t^2},\dfrac{a(1-t^2)}{1+t^2}\right)$ and substituting $(x,y)=(0,0)$, both sides reduce to the same expression $-\dfrac{a(1-t^2)}{1+t^2}$:

\[ -\frac{a(1-t^{2})}{1+t^{2}} \;\stackrel{?}{=}\; \frac{1-t^{2}}{2t}\left(-\frac{2at}{1+t^{2}}\right) = -\frac{a(1-t^{2})}{1+t^{2}}. \tag{3} \]

True for every $t\neq0,\pm1$.

Step 4: Special parameter values

$t=0$: $P=(0,a)$, tangent horizontal, normal $x=0$, through $O$. $t=\pm1$: $P=(\pm a,0)$, tangent vertical, normal $y=0$, through $O$.

Step 5: Verification

\[t=0.6,\ a=1\text{: normal passes through } (0,0) \text{ (numerically, both sides of (3) agree)}\ \checkmark\]