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QJEEM22Mae86e: JEE Main 2022 Mathematics - Class 11 JEEM22Mae86e Solution: For , if is an equilateral triangle with vertices and such that …

For t∈(0,2π)\mathrm{t} \in(0,2 \pi) , if ABC\mathrm{ABC} is an equilateral triangle with vertices A(sin⁡t,−cos⁡t),B(cost⁡,sin⁡t)\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t) and C(a,b)C(a, b) such that its orthocentre lies on a circle with centre (1,13)\left(1, \frac{1}{3}\right) , then (a2−b2)\left(a^{2}-b^{2}\right) is equal to :

Multiple Choice Options

  • A 83\frac{8}{3}
  • B 8
  • C 779\frac{77}{9}
  • D 809\frac{80}{9}

Step-by-Step Verified Solution

Let P(h,k)P(h,k) be the orthocentre of Δ\Delta ABC

JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Circle Question 51 English Explanation

Then

h=sin⁡t+cos⁡t+a3, k=−cos⁡t+sin⁡t+b3h = {{\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{(3h - a)^2} + {(3k - b)^2} = 2

∴\therefore (h−a3)2+(k−b3)2=29{\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}

∵\because Orthocentre lies on circle with centre (1,13)\left( {1,{1 \over 3}} \right)

∴\therefore a=3, b=1a = 3,\,b = 1

∴\therefore a2−b2=8{a^2} - {b^2} = 8

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