Which lines through the origin meet the hyperbola xy = a²

Which lines through the origin meet the hyperbola $xy = a^{2}$ at right angles, that is, are normal to the curve at the point of intersection?

Step 1: The slope of the normal

Differentiating $xy = a^{2}$: $y+xy' = 0$, so $y' = -\dfrac{y}{x}$, and at $P = \bigl(x_{0},y_{0}\bigr)$ with $x_{0}y_{0} = a^{2}$ the normal has slope

\[ k_{n} = -\frac{1}{y'} = \frac{x_{0}}{y_{0}}. \tag{1} \]

Step 2: Passing through the origin

The normal is $y-y_{0} = \dfrac{x_{0}}{y_{0}}\bigl(x-x_{0}\bigr)$; putting $\bigl(x,y\bigr) = \bigl(0,0\bigr)$: \[-y_{0} = -\frac{x_{0}^{2}}{y_{0}} \;\Longrightarrow\; y_{0}^{2} = x_{0}^{2} \;\Longrightarrow\; y_{0} = \pm x_{0}. \tag{2}\]

Step 3: Which sign is possible

Combined with $x_{0}y_{0} = a^{2}\gt 0$: \[y_{0} = x_{0}:\ x_{0}^{2} = a^{2} \;\Longrightarrow\; \bigl(x_{0},y_{0}\bigr) = \bigl(a,a\bigr) \ \text{or} \ \bigl(-a,-a\bigr);\]

\[ y_{0} = -x_{0}:\ -x_{0}^{2} = a^{2}, \ \text{impossible for real } x_{0}. \]

Step 4: The line

At either point $k_{n} = \dfrac{x_{0}}{y_{0}} = 1$, and the line through the origin of slope $1$ is

\[ y = x, \] which indeed contains both $\bigl(a,a\bigr)$ and $\bigl(-a,-a\bigr)$.

Step 5: Verification

\[a\cdot a = a^{2} \;\;\checkmark; \qquad y'\bigl(a,a\bigr) = -\frac{a}{a} = -1,\ \text{so } k_{t}k_{n} = (-1)(1) = -1 \;\;\checkmark\]

\[\text{The two points are the vertices of the hyperbola, nearest to the centre} \;\;\checkmark\]

\[\text{At any other point } \bigl(x_{0},a^{2}/x_{0}\bigr) \text{ the normal misses the origin, since } y_{0}^{2}\neq x_{0}^{2} \;\;\checkmark\]