For t ∈ 0 , 2 π , if A B C is an equilateral triangle with vertices A sin t , - cos t , B cos t ,…

We know that for an equilateral triangle the orthocentre and centroid coincide.

Here, centroid $\left(h,k\right)\equiv \left(\frac{\cos t+\sin t+a}{3},\frac{\sin t-\cos t+b}{3}\right)$

$\Rightarrow 3h-a=\cos t+\sin t ...\left(\mathrm{i}\right)$

$\Rightarrow 3k-b=\sin t-\cos t ...\left(\mathrm{ii}\right)$

Eliminating $t$ from above two equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

${\left(h-\frac{a}{3}\right)}^2+{\left(k-\frac{b}{3}\right)}^2=\frac{2}{9}$

So, $\left(h,k\right)$ lies on the circle whose centre is $\left(\frac{a}{3},\frac{b}{3}\right)$, now comparing with $\left(1,\frac{1}{3}\right)$ we get,$a=3, b=1$

Hence $a^2-b^2=8$