JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 6)
For $$\mathrm{t} \in(0,2 \pi)$$, if $$\mathrm{ABC}$$ is an equilateral triangle with vertices $$\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t)$$ and $$C(a, b)$$ such that its orthocentre lies on a circle with centre $$\left(1, \frac{1}{3}\right)$$, then $$\left(a^{2}-b^{2}\right)$$ is equal to :
$$\frac{8}{3}$$
8
$$\frac{77}{9}$$
$$\frac{80}{9}$$
Explanation
Let $$P(h,k)$$ be the orthocentre of $$\Delta$$ABC
_28th_July_Morning_Shift_en_6_1.png)
Then
$$h = {{\sin t + \cos t + a} \over 3},\,k = {{ - \cos t + \sin t + b} \over 3}$$
(Orthocentre coincide with centroid)
$$\therefore$$ $${(3h - a)^2} + {(3k - b)^2} = 2$$
$$\therefore$$ $${\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}$$
$$\because$$ Orthocentre lies on circle with centre $$\left( {1,{1 \over 3}} \right)$$
$$\therefore$$ $$a = 3,\,b = 1$$
$$\therefore$$ $${a^2} - {b^2} = 8$$
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