Find the locus of the vertex of a right angle whose two

Find the locus of the vertex of a right angle whose two sides are tangent to the hyperbola $b^2x^2-a^2y^2=a^2b^2$.

Step 1: Tangency condition for a line through $P=(h,k)$

A line $y=mx+c$ is tangent to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ exactly when $c^2=a^2m^2-b^2$. If it also passes through $P$, $c=k-mh$, so

\[ (k-mh)^2 = a^2m^2-b^2 \;\Longrightarrow\; m^2(h^2-a^2) - 2khm + (k^2+b^2) = 0. \tag{1} \]

Step 2: The two tangent slopes from $P$

The roots $m_1,m_2$ of (1) are the slopes of the two tangents from $P$; by Vieta, \[m_{1}m_{2} = \frac{k^2+b^2}{h^2-a^2}. \tag{2}\]

Step 3: Perpendicularity

The two tangents meet at a right angle exactly when $m_1m_2=-1$: \[\frac{k^2+b^2}{h^2-a^2} = -1 \;\Longrightarrow\; h^2+k^2 = a^2-b^2. \tag{3}\]

Step 4: Verification

\[\text{Symbolically, } m_1m_2 \text{ from Vieta's formula (2) matches the direct product of the quadratic's roots identically}\ \checkmark\]