A(a, 0) is a fixed point and θ is a parameter such that 0 [AP EAPCET 24 May 2025 S1]

a circle with centre at $\left(\frac{a}{3}, 0\right)$ and radius $\left(\frac{\sqrt{a^2 + b^2}}{3}\right)$
$a$ circle with centre at $(a, 0)$ and radius $\left(\frac{\sqrt{a^2 + b^2}}{3}\right)$
a parabola with focus at $\left(\frac{a}{3}, 0\right)$
a parabola with focus at $(a, 0)$

Solution:

Let the coordinates of the centroid be

G(x, y)

of the

\triangle APQ

x = \frac{x_A + x_P + x_Q}{3}, y = \frac{y_A + y_P + y_Q}{3}

Given the points

A(a, 0), P(a \cos \theta, a \sin \theta)

and

Q(b \sin \theta, -b \cos \theta)

The coordinates of the centroid are

x = \frac{a + a \cos \theta + b \sin \theta}{3}

y = \frac{0 + a \sin \theta - b \cos \theta}{3}

Let

x = h

and

y = k

, then,

h = \frac{a + a \cos \theta + b \sin \theta}{3}, k = \frac{a \sin \theta - b \cos \theta}{3}

\Rightarrow 3h - a = a \cos \theta + b \sin \theta

3k = a \sin \theta - b \cos \theta

Squaring and adding these equations

(3h - a)^2 + (3k)^2 = (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2

= a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \cos \theta \sin \theta

= a^2 + b^2

Replacing

h

with

x

and

k

with

y

, we get the locus of the centroid as

(3x - a)^2 + (3y)^2 = a^2 + b^2

\Rightarrow \left(x - \frac{a}{3}\right)^2 + y^2 = \frac{a^2 + b^2}{9}

\Rightarrow \left(x - \frac{a}{3}\right)^2 + y^2 = \left(\frac{\sqrt{a^2 + b^2}}{3}\right)^2

So, the locus of the centroid is

\left(x - \frac{a}{3}\right)^2 + y^2 = \left(\frac{\sqrt{a^2 + b^2}}{3}\right)^2

It is a circle with centre at

\left(\frac{a}{3}, 0\right)

and radius

\frac{\sqrt{a^2 + b^2}}{3}

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