question:
my attempt:
any $t - $ point on parabola $y^2=2ax$ is given by $\left(\dfrac{at^2}{2}, at\right)$
area of polygon $\ = S=\left|\dfrac{(x_{1}y_{2}-y_{1}x_{2})+(x_{2}y_{3}-y_{2}x_{3})+.................+(x_{n}y_{1}-y_{n}x_{1})}{2}\right|$
where $(x_{i},y_{i})$ are cordinates of $ith$ vertex of polygon
$\implies \ S= \dfrac{a^2}{4}\left(\displaystyle\sum_{cyc} t_{1}t _{2}(t_{1}-t_{2})\right) $
i don't know how to proceed further in order to find cordinates of $(n-1)th$ point on parabola i.e, $A_{n-1} $ ,which isn't vertex of $(n-1)-gon$, such that it's area is maximum
let $A_{n-1}= \left(\dfrac{at^2}{2}, at\right)$;
$A_{n-2}=\left(\dfrac{at_{n-2}^2}{2}, at_{n-2}\right)$
$A_{n}=\left(\dfrac{at_{n}^2}{2}, at_{n}\right)$
by using blue's comment i proceeded as follows:
method 1 : for maximum triangle area tangent at $A_{n-1}$ must be parllel to chord joining $A_{n}A_{n-2}$
slope of tangent at $A_{n}$ = slope of chord $A_{n}A_{n-2}$
$\implies \dfrac{1}{t}=2\left(\dfrac{at_{n-2}-at_{n}}{at_{n-2}^2-at_{n}^2}\right) \implies t= \dfrac{t_{n}+t_{n-2}}{2}$
method 2 :
area of triangle $\Delta= \dfrac{a^2}{4}\left[ t_{n}t_{n-2}(t_{n}-t_{n-2})+t_{n-2}\ t(t_{n-2}-t)+t\ t_{n}(t-t_{n})\right]=f(t)$
for maximising this area
$f'(t)=0\implies t= \dfrac{t_{n}^2-t_{n-2}^2}{2(t_{n}-t_{n-2)}}=\dfrac{t_{n}+t_{n-2}}{2}$
so, cordinates of $A_{n-1}$ will be
$A_{n-1}:\left[\dfrac{a}{2}\left(\dfrac{t_{n}+t_{n-2}}{2}\right)^2,a \left(\dfrac{t_{n}+t_{n-2}}{2}\right)\right]$