For t in (0, 2pi),if ABC is an equilateral tr | vedclass

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(B) Let $s = \sin t$ and $c = \cos t$.For an equilateral triangle,the orthocentre $(h, k)$ coincides with the centroid.The centroid $(h, k)$ is given

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$8$

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(B) Let $s = \sin t$ and $c = \cos t$.For an equilateral triangle,the orthocentre $(h, k)$ coincides with the centroid.The centroid $(h, k)$ is given by $h = \frac{a + s + c}{3}$ and $k = \frac{b - c + s}{3}$.Thus,$3h - a = s + c$ and $3k - b = s - c$.Squaring and adding these equations:$(3h - a)^2 + (3k - b)^2 = (s + c)^2 + (s - c)^2 = 2(s^2 + c^2) = 2$.Dividing by $9$:$(h - a/3)^2 + (k - b/3)^2 = 2/9$.This represents a circle with centre $(a/3, b/3)$ and radius $\sqrt{2}/3$.Given the centre of the circle is $(1, 1/3)$,we have $a/3 = 1$ and $b/3 = 1/3$.Therefore,$a = 3$ and $b = 1$.Finally,$a^2 - b^2 = 3^2 - 1^2 = 9 - 1 = 8$.

For $t \in (0, 2\pi)$,if $ABC$ is an equilateral triangle with vertices $A(\sin t, -\cos t)$,$B(\cos t, \sin t)$,and $C(a, b)$ such that its orthocentre lies on a circle with centre $(1, 1/3)$,then $(a^2 - b^2)$ is equal to.

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