6. The graph of $x^{2}-(y-1)^{2}=1$ has one tangent line with positive slope that passes through $(x, y)=(0,0)$. If the point of tangency is $(a, b)$, find $\sin ^{-1}\left(\frac{a}{b}\right)$ in radians.
Solution: Differentiating both sides of the equation, we find that $2 x-2(y-1) \frac{\mathrm{d} y}{\mathrm{~d} x}=0$, and so $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{y-1}=\frac{a}{b-1}$. The line passing through $(0,0)$ and $(a, b)$ has slope $\frac{b}{a}$, so $\frac{b}{a}=\frac{a}{b-1}$. Solving simultaneously with $a^{2}-(b-1)^{2}=1$, we get $b^{2}-b-\left[(b-1)^{2}+1\right]=0$, and so $b=2, a=\sqrt{(2)}$. Finally, $\sin ^{-1}\left(\frac{a}{b}\right)=\frac{\pi}{4}$.