Hello, Blasnmt23!
I used a simpler approach . . .
Code:
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o | ....♥ B
| ....*::::::*
....*::::::::::::*
....*::::::|:::::::::::*
A ♥:::::::::::::|::::::::::* o
*:::::::::|:::::::::*
o *:::::|::::::::* o
o *:|:::::::* o
o | *::::* o
o | ♥
- - - - - - - - o - - C - - - - - - -
|
Point 
has coordinates )
Point 
has coordinates )
Point 
has coordinates )
The area of 
is given by: . 
We have: . ![A \;=\;\tfrac{1}{2}\bigg[(bx^2 - b^2x) - (ax^2 -a^2x) + (ab^2-a^2b)\bigg]](http://latex.codecogs.com/png.latex?A \;=\;\tfrac{1}{2}\bigg[(bx^2 - b^2x) - (ax^2 -a^2x) + (ab^2-a^2b)\bigg])
. . . . . . . . ![A \;=\;\tfrac{1}{2}\bigg[(b-a)x^2 - (b^2-a^2)x + ab(b-a)\bigg]](http://latex.codecogs.com/png.latex?A \;=\;\tfrac{1}{2}\bigg[(b-a)x^2 - (b^2-a^2)x + ab(b-a)\bigg])
Maximize: . ![\frac{dA}{dx} \;=\;\tfrac{1}{2}\bigg[2(b-a)x - (b^2-a^2)\bigg] \;=\;0](http://latex.codecogs.com/png.latex?\frac{dA}{dx} \;=\;\tfrac{1}{2}\bigg[2(b-a)x - (b^2-a^2)\bigg] \;=\;0 )
. . . . . . . x \:=\:b^2-a^2 \quad\Rightarrow\quad x \;=\;\frac{(b-a)(b+a)}{2(b-a)} \quad\Rightarrow\quad x \;=\;\frac{a+b}{2})
The coordinates of point are: . ![\left(\frac{a+b}{2},\;\left[\frac{a+b}{2}\right]^2\right)](http://latex.codecogs.com/png.latex?\left(\frac{a+b}{2},\;\left[\frac{a+b}{2}\right]^2\right) )
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Fascinating!
Point is exactly "halfway between" points and 
Moreover, is the point at which the tangent is parallel to the secant 
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Maximum Area
intersects the parabola
in points A and B.
