SOLUTION TO PROBLEM H-3

We begin by assuming that P lies in the right-half-plane, that P=(a,a³) for some a>0. (If P=(0,0) the tangent will not intersect the curve again.) The tangent line at P is given by

y=3a²x-2a³

To find x coordinate of Q, we need to solve x³=3a²x-2a³ or x³-3a²x+2a³=0. Since x=a satisfies this cubic, x-a is a factor, and

0=x³-3a²x+2a³=(x-a)(x²+ax-2a²)=(x-a)²(x+2a)

So Q=(-2a,-8a³). The tangent line at Q is given by

y=12a²x+16a³

Again we need to solve a cubic, x³=12a²x+16a³. Again we know a root, namely x=-2a. So the tangent line at Q intersects y=x³ at (4a, 64a³). As a result,

$$\begin{eqnarray*}A&=&\int_{-2a}^a x^3-(3a^2x-2a^3)\; dx = 27a^4/4\\ B&=&\int_{-2a}^{4a} (12a^2x+16a^3)-x^3\; dx = 108a^4 \end{eqnarray*}$$

So

$$\frac{B}{A}=16$$

Because y=x³ is symmetric about the origin, this relationship holds for P in the left-half-plane as well.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Feb 1 18:59:37 CST 2002