If the locus of the centroid of the triangle | vedclass

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(B) Let the centroid be $(x, y)$. The coordinates of the centroid are given by: $(x, y) = \left(\frac{a + a \cos t + b \sin t}{3}, \frac{0 + a \sin t

Vedclass · Accepted Answer

$\frac{7}{2}$

Vedclass · Answer

(B) Let the centroid be $(x, y)$. The coordinates of the centroid are given by: $(x, y) = \left(\frac{a + a \cos t + b \sin t}{3}, \frac{0 + a \sin t - b \cos t}{3}ight)$ This implies: $3x = a + a \cos t + b \sin t \Rightarrow 3x - a = a \cos t + b \sin t$ $3y = a \sin t - b \cos t$ Squaring and adding these two equations: $(3x - a)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$ $9x^2 + a^2 - 6ax + 9y^2 = a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t)$ $9x^2 + 9y^2 - 6ax + a^2 = a^2 + b^2$ $9x^2 + 9y^2 - 6ax = b^2$ Comparing this with the given locus $9x^2 + 9y^2 - 6x = 49$,we get $a = 1$ and $b^2 = 49$,so $b = 7$. The line equation is $\frac{x}{1} + \frac{y}{7} = 1$. The intercepts are $x = 1$ and $y = 7$. The area of the triangle formed with the coordinate axes is $\frac{1}{2} 	imes |x_{intercept}| 	imes |y_{intercept}| = \frac{1}{2} 	imes 1 	imes 7 = \frac{7}{2}$.

If the locus of the centroid of the triangle with vertices $A(a, 0)$,$B(a \cos t, a \sin t)$ and $C(b \sin t, -b \cos t)$ ($t$ is a parameter) is $9x^2 + 9y^2 - 6x = 49$,then the area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

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