Quadrature of Parabola

Theorem

Let $T$ be a parabola.

Consider the parabolic segment bounded by an arbitrary chord $AB$.

Let $C$ be the point on $T$ where the tangent to $T$ is parallel to $AB$.

Let

Then the area $S$ of the parabolic segment $ABC$ of $T$ is given by:

$S = \frac 4 3 \triangle ABC$

Proof

We consider WLOG the parabola $y = a x^2$.

Let $A, B, C$ be the points:

• $A = \left({x_0, a x_0^2}\right)$
• $B = \left({x_2, a x_2^2}\right)$
• $C = \left({x_1, a x_1^2}\right)$

The slope of the tangent at $C$ is given by using:

$\frac{\mathrm{d}{y}}{\mathrm{d}{x}} 2 a x_1$

which is parallel to $AB$.

Thus:

$2 a x_1 = \frac {ax_0^2 - a x_2^2} {x_0 - x_2}$

which leads to

$x_1 = \frac {x_0 + x_2} 2$

So the vertical line through $C$ is a bisector of $AB$, at point $P$.

Now, complete the parallelogram $CPBQ$.

Also, find $E$ which is the point where the tangent to $T$ is parallel to $BC$.

By the same reasoning, the vertical line through $E$ is a bisector of $BC$, and so it also bisects $BP$ at $H$.

Next:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle EF\)	\(=\)	\(\displaystyle a \left({\frac {x_1 + x_2} 2}\right)^2 - \left({a x_1^2 + 2 a x_1 \frac {x_2 - x_1} 2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a 4 \left({\left({x_1 + x_2}\right)^2 - 4 x_1^2 + 4 x_1 \left({x_2 - x_1}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a 4 \left({x_1^2 - 2 x_1 x_2 + x_2^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a 4 \left({x_2 - x_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

At the same time:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle QB\)	\(=\)	\(\displaystyle a x_2^2 - \left({a x_1^2 + 2 a x_1 \left({x_2 - x_2}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \left({x_1^2 - 2 x_1 x_2 + x_2^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \left({x_2 - x_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $QB = 4 FE = FH$ and because $CB$ is the diagonal of a parallelogram, $2 FE = 2 EG = FG$.

This implies that $2 \triangle BEG = \triangle BGH$ and $2 \triangle CEG = \triangle BGH$

So $\triangle BCE = \triangle BGH$ and so as $\triangle BCP = 4 \triangle BGH$ we have that:

$BCE = \frac {\triangle BCP} 4$

A similar relation holds for $\triangle APC$:

... so it can be seen that $\triangle ABC = 4 \left({\triangle ADC + \triangle CEB}\right)$.

Similarly, we can create four more triangles underneath $\triangle ADC$ and $\triangle CEB$ which are $\frac 1 4$ the area of those combined, or $\frac 1 {4^2} \triangle ABC$.

This process can continue indefinitely.

So the area $S$ is given as:

$S = \triangle ABC \left({1 + \frac 1 4 + \frac 1 {4^2} + \cdots}\right)$

But from Sum of Geometric Progression it follows that:

$S = \triangle ABC \left({\frac 1 {1 - \frac 1 4}}\right) = \frac 4 3 \triangle ABC$

$\blacksquare$

Historical Note

This proof was given by Archimedes in his book Quadrature of the Parabola, except that he used a different technique to prove that $\triangle ADC + \triangle CEB = \frac {\triangle ABC} 4$.

Sources

• George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.3$