1. ## Maximum Area

Thanks!

2. Hello, Blasnmt23!

I used a simpler approach . . .

The line $y \:=\: mx + b$ intersects the parabola $y = x^2$ in points A and B.
Find the point $C$ that maximizes the area of the triangle.
Code:
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o              |          ....♥ B
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A ♥:::::::::::::|::::::::::*  o
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o      *:::::|::::::::*   o
o         *:|:::::::*   o
o         | *::::*  o
o     |     ♥
- - - - - - - - o - - C - - - - - - -
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Point $A$ has coordinates $(a,a^2)$
Point $B$ has coordinates $(b,b^2)$
Point $C$ has coordinates $(x,x^2)$

The area of $\Delta ABC$ is given by: . $A \;=\;\frac{1}{2}\left|\begin{array}{ccc}1&1&1 \\ a&b&x \\ a^2&b^2&x^2\end{array}\right|$

We have: . $A \;=\;\tfrac{1}{2}\bigg[(bx^2 - b^2x) - (ax^2 -a^2x) + (ab^2-a^2b)\bigg]$

. . . . . . . . $A \;=\;\tfrac{1}{2}\bigg[(b-a)x^2 - (b^2-a^2)x + ab(b-a)\bigg]$

Maximize: . $\frac{dA}{dx} \;=\;\tfrac{1}{2}\bigg[2(b-a)x - (b^2-a^2)\bigg] \;=\;0$

. . . . . . . $2(b-a)x \:=\:b^2-a^2 \quad\Rightarrow\quad x \;=\;\frac{(b-a)(b+a)}{2(b-a)} \quad\Rightarrow\quad x \;=\;\frac{a+b}{2}$

The coordinates of point $C$ are: . $\left(\frac{a+b}{2},\;\left[\frac{a+b}{2}\right]^2\right)$

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Fascinating!

Point $C$ is exactly "halfway between" points $A$ and $B.$

Moreover, $C$ is the point at which the tangent is parallel to the secant $AB.$