MHB Find maximum area of a triangle

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $\triangle ABC$.
 
Mathematics news on Phys.org
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]
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top