Quadrature of Parabola/Proof 2

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 $AB$.

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

$S = \dfrac 4 3 \triangle ABC$

Proof

Let $T$ be the parabola which is the locus of points $\tuple {x, y}$ satisfying $y = x^2$.

By Area of Triangle Inscribed in Parabola:

the point $C$ where the tangent to $T$ at $C$ is parallel to $AB$ has $x$-coordinate $\dfrac 1 2 \paren {x_0 + x_2}$.

Let $d$ be the horizontal distance between $A$ and $B$.

By Area of Triangle Inscribed in Parabola:

$\map \AA {\triangle ABC} = \dfrac 1 8 d^3$

Now construct two triangles.

One is $BC$ bisected with point $E$ on the parabola, forming:

$\triangle EBC$

The other is $AC$ bisected with point $D$ on the parabola, forming:

$\triangle DAC$

For each triangle, the length along the $x$-coordinate $\Delta x$ is equal to $\dfrac d 2$.

The area of each triangle is then:

$\AA = \dfrac 1 8 \paren {\dfrac d 2}^3 = \dfrac 1 8 \cdot \dfrac 1 8 d^3$

From above:

$\map \AA {\triangle ABC} = \dfrac 1 8 d^3$

Thus

$\map \AA {\triangle EBC} = \map \AA {\triangle DAC} = \dfrac 1 8 \map \AA {ABC}$

For two triangles, the area together is:

$\map \AA {\triangle DAC} + \map \AA {\triangle EBC} = \dfrac 1 4 \map \AA {ABC}$

By a variant of the method of exhaustion, the total area over the parabola and under $AB$ is:

$\map \AA {ABC} \cdot \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \dots}$

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By Sum of Infinite Geometric Sequence:

$1 + \dfrac 1 4 + \dfrac 1 {4^2} + \dots = \dfrac 4 3$

Hence the total area is:

$S = \map \AA {ABC} \cdot \dfrac 4 3$

$\blacksquare$