Parabolic Hull

Without loss of generality, let us assume $f(x)$ is in the form $y=ax^2$ . Let $P(p,ap^2)$ be a variable point on the parabola, and $Q(q,aq^2)$ be a point such that the area between PQ and the parabola is the constant $A$ .

It is easy to show that the equation of PQ is $y=a(p+q)x-apq$ , and so we can find the area by integrating the difference between PQ and $f(x)$ : $\begin{aligned}\int_p^q a(p+q)x-apq-ax^2\ dx&=a\int_p^q (p+q)x-pq-x^2\ dx\\ &=a\left[\frac{(p+q)x^2}2-pqx-\frac{x^3}3\right]_p^q\\ &=a\left(\frac{q^3+pq^2}2-pq^2-\frac{q^3}3\right)-a\left(\frac{p^3+p^2q}2-p^2q-\frac{p^3}3\right)\\ &=\frac a6\left(3q^3+3pq^2-6pq^2-2q^3-3p^3-3p^2q+6p^2q+2p^3\right)\\ &=\frac a6\left(q^3-3pq^2+3p^2q-p^3\right)\\ &=\frac {a(q-p)^3}6\end{aligned}$ Since this area must equal $A$ , we can deduce that $q=p+\sqrt[3]\frac{6A}a$ . For simplicity, let $k=\sqrt[3]\frac{6A}a$ .

Now the equation for each secant depends on the single variable $p$ : $y=a(2p+k)x-ap(p+k)$ .

Consider 2 more secants, which are drawn from points on the parabola with x-coordinates $p+h$ and $p-h$ , and cutting off the same area as above. The former would have equation $y=a[2(p+h)+k]x-a(p+h)(p+h+k)$ . Solving simultaneously with the equation of PQ, we can find the intersection point between PQ and the $p+h$ secant: $\begin{aligned}a(2p+2h+k)x-a(p+h)(p+h+k)&=a(2p+k)x-ap(p+k)\\ [a(2p+2h+k)-a(2p+k)]x&=a(p+h)(p+h+k)-ap(p+k)\\ x&=\frac{(p+h)(p+h+k)-p(p+k)}{(2p+2h+k)-(2p+k)}\\ x&=\frac{p^2+2ph+h^2+ph+hk-p^2-ph}{2p+2h+k-2p-k}\\ x&=\frac{2ph+hk+h^2}{2h}\\ x&=\frac{2p+k+h}{2}\end{aligned}$ Similarly it can be shown that the x-coordinate of the intersection between PQ and the $p-h$ secant is $\frac{2p+k-h}{2}$ .

Now, $\lim_{h\to0}\frac{2p+k+h}{2}=\lim_{h\to0}\frac{2p+k-h}{2}=\frac{2p+k}{2}$ so the intersection points approach each other as the secants approach each other.

Now, if we draw a curve that is tangent to all three secants, the tangent point on PQ must be between the two intersection points, as shown in the following diagram. But since the two intersection points approach the same point $x=\frac{2p+k}{2}$ as the difference between the secants approaches zero, by squeeze theorem we can conclude that the tangent point must be at $x=\frac{2p+k}{2}$ as the number of secants approaches infinity. Therefore $g(x)$ must be the locus of this intersection point as $p$ varies.

The y-coordinate can be found by substituting x into PQ, and we can find the locus by eliminating $p$ : $\begin{aligned}x&=\frac{2p+k}{2}&y&=a(2p+k)x-ap(p+k)\\ \therefore 2p+k&=2x&&=a(2x)x-a\left(x-\frac{k}{2}\right)\left(x-\frac{k}{2}+k\right)\\ \therefore p&=x-\frac{k}{2}&&=2ax^2-a\left(x-\frac{k}{2}\right)\left(x+\frac{k}{2}\right)\\ &&&=2ax^2-ax^2+\frac{ak^2}{4}\\ &&y&=ax^2+\frac{ak^2}{4}\end{aligned}$

We can see that $g(x)$ is the same as $f(x)$ , but moved up by $\frac{ak^2}{4}$ units. Therefore the parabolas are congruent, and have the same focal length .