# Quadrature of Parabola

## Contents

## 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:

- $S = \dfrac 4 3 \triangle ABC$

## Proof

Without loss of generality, consider the parabola $y = a x^2$.

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

\(\displaystyle A\) | \(=\) | \(\displaystyle \tuple {x_0, a {x_0}^2}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle B\) | \(=\) | \(\displaystyle \tuple {x_2, a {x_2}^2}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle C\) | \(=\) | \(\displaystyle \tuple {x_1, a {x_1}^2}\) | $\quad$ | $\quad$ |

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

- $\dfrac {\d y} {\d x} 2 a x_1$

which is parallel to $AB$.

Thus:

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

which leads to:

- $x_1 = \dfrac {x_0 + x_2} 2$

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

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:

\(\displaystyle EF\) | \(=\) | \(\displaystyle a \paren {\frac {x_1 + x_2} 2}^2 - \paren {a x_1^2 + 2 a x_1 \frac {x_2 - x_1} 2}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac a 4 \paren {\paren {x_1 + x_2}^2 - 4 {x_1}^2 + 4 x_1 \paren {x_2 - x_1} }\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac a 4 \paren { {x_1}^2 - 2 x_1 x_2 + {x_2}^2}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac a 4 \paren {x_2 - x_1}^2\) | $\quad$ | $\quad$ |

At the same time:

\(\displaystyle QB\) | \(=\) | \(\displaystyle a {x_2}^2 - \paren {a {x_1}^2 + 2 a x_1 \paren {x_2 - x_2} }\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \paren { {x_1}^2 - 2 x_1 x_2 + {x_2}^2}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \paren {x_2 - x_1}^2\) | $\quad$ | $\quad$ |

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:

- $BCE = \dfrac {\triangle BCP} 4$

A similar relation holds for $\triangle APC$:

so it can be seen that:

- $\triangle ABC = 4 \paren {\triangle ADC + \triangle CEB}$

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

This process can continue indefinitely.

So the area $S$ is given as:

- $S = \triangle ABC \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \cdots}$

But from Sum of Geometric Progression it follows that:

- $S = \triangle ABC \paren {\dfrac 1 {1 - \dfrac 1 4} } = \dfrac 4 3 \triangle ABC$

$\blacksquare$

## Historical Note

The quadrature of a parabola was first given by Archimedes in his book *On the Quadrature of the Parabola*.

His proof was the same as the one documented here, except that he used a different technique to prove:

- $\triangle ADC + \triangle CEB = \dfrac {\triangle ABC} 4$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 4$: Geometric Formulas: $4.24$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.3$: Archimedes' Quadrature of the Parabola