For $\mathrm{t} \in(0,2 \pi)$, if ABC is an equilateral triangle with vertices $\mathrm{A}(\sin t,-\cos t), \mathrm{B}(\cos t, \sin t)$ and $\mathrm{C}(\mathrm{a}, \mathrm{b})$ 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 :

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Straight Lines

For

t \in (0, 2 π)

, 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, \frac{1}{3})

, then

(a^{2} - b^{2})

is equal to :

Answer: (B)

For an equilateral triangle, the orthocentre coincides with the centroid. Let the orthocentre be

(h, k)

. Then

h = \frac{s i n t + c o s t + a}{3}

and

k = \frac{- c o s t + s i n t + b}{3}

. This gives

a + sin t + cos t = 3 h

and

b + sin t - cos t = 3 k

. The distance from A to the centroid squared is

(sin t - h)^{2} + (- cos t - k)^{2} = (\frac{c o s t + a}{3})^{2} + (\frac{s i n t - c o s t + b}{3})^{2}

. After simplification, we get

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

, which is a circle centered at

(\frac{a}{3}, \frac{b}{3})

. Comparing with the given center

(1, \frac{1}{3})

\frac{a}{3} = 1

and

\frac{b}{3} = \frac{1}{3}

, giving

a = 3

and

b = 1

. Therefore

a^{2} - b^{2} = 9 - 1 = 8

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option