A parabola y² = 2px intersects a circle x² + y² - 2rx = 0

A parabola $y^2 = 2px$ intersects a circle $x^2 + y^2 - 2rx = 0$ at points $A$ and $B$. Determine the parameter $p$ such that the area bounded by the parabola and the chord $AB$ is maximized.

We find the intersection points, build the area as a function of $p$, and maximize it.

Step 1: Rewrite the circle

\[x^2-2rx+y^2=0 \implies (x-r)^2+y^2=r^2\]

a circle centered at $(r,0)$ with radius $r$.

Step 2: Find the intersection points

Substituting $y^2=2px$ into the circle's equation:

\[x^2+2px-2rx=0 \implies x[x+2(p-r)]=0\]

So $x=0$ (the origin) or $x=2(r-p)$. For $x=2(r-p)$: $y^2=2p\cdot2(r-p)=4p(r-p)$, so $y=\pm2\sqrt{p(r-p)}$ (real provided $0 < p < r$).

The chord $AB$ is the vertical segment at $x=2(r-p)$ between $A=\left(2(r-p),\,2\sqrt{p(r-p)}\right)$ and $B=\left(2(r-p),\,-2\sqrt{p(r-p)}\right)$.

Step 3: Set up the area as a function of $p$

Integrating with respect to $y$ (using the symmetry about the $x$-axis), with the parabola $x=\frac{y^2}{2p}$ and the chord at $x=2(r-p)$:

\[A(p) = 2\int_0^{2\sqrt{p(r-p)}}\left[2(r-p)-\frac{y^2}{2p}\right]dy = 2\left[2(r-p)y-\frac{y^3}{6p}\right]_0^{2\sqrt{p(r-p)}}\]

\[= 2\left[4(r-p)\sqrt{p(r-p)}-\frac{4(r-p)\sqrt{p(r-p)}}{3}\right] = \frac{16}{3}(r-p)\sqrt{p(r-p)}\]

Step 4: Reduce to maximizing a polynomial

Since $A(p)=\frac{16}{3}\sqrt{p}\,(r-p)^{3/2} = \frac{16}{3}\sqrt{p(r-p)^3}$, maximizing $A(p)$ is equivalent to maximizing $f(p)=p(r-p)^3$ on $0 < p < r$.

Step 5: Find the critical point

\[f'(p) = (r-p)^3 - 3p(r-p)^2 = (r-p)^2\left[(r-p)-3p\right] = (r-p)^2(r-4p)\]

Setting $f'(p)=0$ with $0 < p < r$ (so $(r-p)^2\ne0$): $r-4p=0 \implies p=\dfrac{r}{4}$.

Step 6: Verify this is a maximum

For $p<\frac r4$: $r-4p>0$, so $f'(p)>0$ (increasing). For $p>\frac r4$: $r-4p<0$, so $f'(p)<0$ (decreasing). Hence $p=\frac r4$ gives a maximum. ✓

\[\boxed{p = \frac{r}{4}}\]

Summary

\[\boxed{p = \frac{r}{4}}\]

Key insight: The enclosed area $A(p)=\frac{16}{3}\sqrt{p(r-p)^3}$ vanishes at both endpoints $p=0$ and $p=r$ (the chord degenerates to a point in each case), so a genuine interior maximum must exist; it occurs at the one-quarter point $p=\frac r4$, independent of the specific value of $r$.