Find maximum area of a triangle

[anemone]

Given two moving points $A(x_1,\,y_1)$ and $B(x_2,\,y_2)$ on parabola curve $y^2=6x$ with $x_1+x_2=4$ and $x_1\ne x_2$ and the perpendicular bisector of segment $AB$ intersects $x$-axis at point $C$. Find the maximum area of $\triangle ABC$.

 

[Opalg]

Spoiler

This seems to be one of those problems where the calculation is easier if you make it more general. I shall take the parabola to be $y^2 = 4ax$ (so eventually $a$ will become $\frac32$), and the condition on the points to be $x_1+x_2 = k$ (with $k=4$ eventually).

Let $A$ and $B$ be the points $(as^2, 2as)$, $(at^2,2at)$, so that $as^2 + at^2 = k$. The line $AB$ has slope $\dfrac2{s+t}$ and midpoint $\bigl(\frac12k,a(s+t)\bigr)$. Its perpendicular bisector therefore has equation $y - a(s+t) = -\tfrac12(s+t)(x - \tfrac12k)$, and it meets the $x$-axis at the point $(2a+\tfrac12k,0)$. The area $\Delta$ of triangle $ABC$ is the absolute value of $$\frac12\begin{vmatrix}as^2&2as&1\\ at^2&2at&1\\ 2a+\frac12k&0&1 \end{vmatrix} = (2a+\tfrac12k)a(s-t) + a^2st(s-t).$$ Now let $u = s-t$ and $v = s+t$. Then $u^2+v^2 = 2(s^2+t^2) = \frac{2k}a$ and $v^2-u^2 = 4st$, so that $st = \frac14(v^2-u^2) = \frac12\bigl(\frac ka - u^2\bigr)$. Therefore $$\Delta = \left| (2a+\tfrac12k)a(s-t) + a^2st(s-t) \right| = \left| (2a+\tfrac12k)au + a^2\tfrac12\bigl(\tfrac ka - u^2\bigr)u\right| = \left| (2a+k)au - \tfrac12a^2u^3 \right|.$$ Now plug in the values $a=\frac32$ and $k=4$ to get $\Delta = \left| \frac{21}2u - \frac98u^3\right|$. Also, $u^2 \leqslant u^2+v^2 = \frac{2k}a = \frac{16}3$, so that $u$ must lie in the interval $\left[-\frac4{\sqrt3},\frac4{\sqrt3}\right]$. The maximum value of $ \left| \frac{21}2u - \frac98u^3\right|$ in that interval occurs at the turning point when $\frac{21}2 - \frac{27}8u^2 = 0$ and so $u = \frac{2\sqrt7}3$. Therefore the maximum value of $\Delta$ is $\Delta_{\max} = \frac{2\sqrt7}3\left(\frac{21}2 - \frac98\frac{28}9\right) = \frac{14}3\sqrt7 \approx 12.347$.

Having found $u$, it is easy to work out corresponding values of $s$ and $t$, which can be taken as $\frac13(\sqrt7\pm\sqrt5)$. The points $A$, $B$ and $C$ are then $$A = \left( 2 - \tfrac13\sqrt{35}, \sqrt7 - \sqrt5\right) \approx (0.08,0.41),$$ $$B = \left( 2 + \tfrac13\sqrt{35}, \sqrt7 + \sqrt5\right) \approx (3.97,4.88),$$ $$C = (5,0).$$
[TIKZ]\draw [help lines, ->] (-1,0) -- (6.5,0) ;
\draw [help lines, ->] (0,-3) -- (0,6.5) ;
\draw [domain=-1:2, samples=100] plot ({1.5*\x*\x}, {3*\x });
\coordinate [label=left:$A$] (A) at (0.03,0.41) ;
\coordinate [label=above left:$B$] (B) at (3.97,4.88) ;
\coordinate [label=below:$C$] (C) at (5,0) ;
\draw [thick] (A) -- (B) -- (C) -- cycle ;
\draw (2,2.65) -- (C) ;
\foreach \x in {2,4,6} { \draw (\x,-0.1) node [ below ] {$\x$} -- (\x,0.1) ;
\draw (-0.1,\x) node [ left ] {$\x$} -- (0.1,\x) ; } ;[/TIKZ]