Quadrature of Parabola

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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 $AB$.


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

$S = \dfrac 4 3 \triangle ABC$


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 $AB$.


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

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 $H$.


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

At the same time:

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


$QB = 4 FE = FH$

and because $CB$ is the diagonal of a parallelogram:

$2 FE = 2 EG = FG$

This implies that:

$2 \triangle BEG = \triangle BGH$


$2 \triangle CEG = \triangle BGH$


$\triangle BCE = \triangle BGH$

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:

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


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:

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