Area of Triangle Inscribed in Parabola/Proof 2
Theorem
Let $T$ be the parabola which is the locus of points $\tuple {x, y}$ satisfying $y = x^2$.
Let $A$ and $B$ be two arbitrary points on $T$ with coordinates:
- $A = \tuple {u, u^2}$
- $B = \tuple {v, v^2}$, with $u > v$
Let $C$ be a third point on $T$ whose $x$-coordinate is the average of those of $A$ and $B$.
The area of $\triangle ABC$ is:
- $\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$
Proof
From Equation of Straight Line in Plane through Two Points, the straight line $AB$ can be expressed as:
| \(\ds \dfrac {\paren {y_{AB} - y_1} } {\paren {x - x_1} }\) | \(=\) | \(\ds \dfrac {\paren {y_2 - y_1} } {\paren {x_2 - x_1} }\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\paren {y_{AB} - v^2} } {\paren {x - v} }\) | \(=\) | \(\ds \dfrac {\paren {u^2 - v^2} } {\paren {u - v} }\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\paren {u + v} \paren {u - v} } {\paren {u - v} }\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {u + v}\) | dividing top and bottom by $\paren {u - v}$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {y_{AB} - v^2}\) | \(=\) | \(\ds \paren {u + v} \paren {x - v}\) | multiplying both sides by $\paren {x - v}$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {u + v} x - u v - v^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y_{AB}\) | \(=\) | \(\ds \paren {u + v} x - u v\) |
The coordinates of $C$ are:
- $\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $
Let $P$ be the midpoint of $AB$, such that $AP = BP$.
We can use Equation of Straight Line in Plane through Two Points above to determine the coordinates of $P$:
| \(\ds y_{AB}\) | \(=\) | \(\ds \paren {u + v} x - u v\) | from above | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {u + v} \paren {\dfrac 1 2 \paren {u + v} } - u v\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac 1 2 \paren {u^2 + v^2}\) |
Therefore, the coordinates of $P$ are:
- $\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 2 \paren {u^2 + v^2} } $
We now connect points $P$ and $C$ to create line segment $PC$
This in turn creates two triangles.
We see from the diagram above:
- $\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle CPB}$
From Area of Triangle in Terms of Two Sides and Angle, we have:
- $\map \AA {\triangle APC} = \dfrac 1 2 \paren {AP} \paren {PC} \map \sin {\angle APC} $
We now notice that:
| \(\ds \paren {AP} \map \sin {\angle APC}\) | \(=\) | \(\ds \dfrac {\paren {u + v} } 2 - v\) | from above | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\paren {u - v} } 2\) |
Therefore:
- $\map \AA {\triangle APC} = \dfrac 1 2 \paren {PC} \dfrac {\paren {u - v} } 2$
We also notice:
| \(\ds \map \sin {\angle CPB}\) | \(=\) | \(\ds \map \sin {\pi - \angle APC}\) | $\angle CPB$ is supplementary to $\angle APC$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sin {\angle APC}\) | Sine of Supplementary Angle |
Since $AP = PB$, $PC = PC$ and $\map \sin {\angle CPB} = \map \sin {\angle APC}$, then by Area of Triangle in Terms of Two Sides and Angle we have:
- $\map \AA {\triangle APC} = \map \AA {\triangle CPB}$
Therefore:
| \(\ds \map \AA {\triangle ABC}\) | \(=\) | \(\ds \map \AA {\triangle APC} + \map \AA {\triangle CPB}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \AA {\triangle APC}\) | since $\map \AA {\triangle APC} = \map \AA {\triangle CPB}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \dfrac 1 2 \paren {PC} \dfrac {\paren {u - v} } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {u - v} P C\) |
All that remains is to determine the length of $CP$:
| \(\ds CP\) | \(=\) | \(\ds P_y - C_y\) | Definition of Length of Line | |||||||||||
| \(\ds CP\) | \(=\) | \(\ds \dfrac 1 2 \paren {u^2 + v^2} - \dfrac 1 4 \paren {u + v}^2\) | substitution | |||||||||||
| \(\ds CP\) | \(=\) | \(\ds \dfrac 1 4 \paren {2 u^2 + 2 v^2 - \paren {u + v}^2}\) | factor out $\dfrac 1 2$ | |||||||||||
| \(\ds CP\) | \(=\) | \(\ds \dfrac 1 4 \paren {u^2 - 2 u v + v^2}\) | expansion and subtraction | |||||||||||
| \(\ds CP\) | \(=\) | \(\ds \dfrac 1 4 \paren {u - v}^2\) | factor |
Substituting from above:
- $\map \AA {\triangle ABC} = \dfrac 1 2 \paren {u - v} \dfrac 1 4 \paren {u - v}^2$
Therefore:
- $\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$
$\blacksquare$