For t ∈ (0, 2π), if ABCis an equilateral triangle with vertices A(...

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

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

$⇒ 3h - a = \cos t + \sin t . . . \left(i\right)$

$⇒ 3k - b = \sin t - \cos t . . . \left(ii\right)$

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

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

So, \left(h, k\right)$lies on the circle whose centre is \left(\dfrac{a}{3}, \dfrac{b}{3}\right)$, now comparing with \left(1, \dfrac{1}{3}\right)$we get,$a = 3, b = 1$</p><p>Hence${a}^{2} - {b}^{2} = 8