Mathematical Immaturity

14.15 Exercises

• For certain values of the constants $a$ and $b,$ the two curves with Cartesian equations $y = ax(b - x)$ and $(x + 2)y = x$ intersect at only one point $P,$ have a common tangent line at $P,$ and have the same curvature at $P.$ (a) Find all $a$ and $b$ which satisfy all these conditions. (b) For each possible choice of $a$ and $b$ satisfying the given conditions, make a sketch of the two curves. Show how they intersect at $P.$
• What happens if we set $x = 0\,?$ What about $y = 0\,?$
• Setting $x = 0$ or $y = 0,$ we see that both equations become zero. And since there is exactly one point $P$ at which the two curves intersect, then $P = (0, 0).$ Therefore, we wish to find $a$ and $b$ such that at $P = (0, 0),$ the derivatives of both equations with respect to $x$ are equal. Additionally, following the result of Exercise 8, we wish to find $a$ and $b$ such that the absolute values of the curves' second derivatives with respect to $x$ are also equal at $P = (0, 0).$ In other words, we wish to find $a$ and $b$ such that when $x = 0$ $$ \begin{align*} \\ a(b - 2x) &= \frac{1}{x + 2} + \frac{x}{(x + 2)^2} \\ \\ |a| &= \frac{x}{(x + 2)^3} - \frac{1}{(x + 2)^2} \end{align*} $$ Plugging in $x = 0,$ these equations simplify to $$ \begin{align*} \\ ab &= \frac{1}{2}; \quad |a| = \frac{1}{4} \end{align*} $$ This is satisfied either when $a = -1/4$ and $b = -2$ or when $a = 1/4$ and $b = 2. \quad \blacksquare$