Let P(3 cos alpha, 2 sin alpha), alpha neq 0, | vedclass

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(D) Let $Q(x_Q, y_Q)$ and $R(x_R, y_R)$. The centroid formula for triangle $PQR$ is given by $(\frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3})$.

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $Q(x_Q, y_Q)$ and $R(x_R, y_R)$. The centroid formula for triangle $PQR$ is given by $(\frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3})$.Given the centroid is $(2 + \cos \alpha, 3 + \frac{2}{3} \sin \alpha)$, we have:$\frac{3 \cos \alpha + x_Q + x_R}{3} = 2 + \cos \alpha \implies x_Q + x_R = 6$.$\frac{2 \sin \alpha + y_Q + y_R}{3} = 3 + \frac{2}{3} \sin \alpha \implies y_Q + y_R = 9$.Point $Q$ lies on the circle $x^2 + y^2 - 14x - 14y + 82 = 0$, which can be written as $(x-7)^2 + (y-7)^2 = 16$.Point $R$ lies on the line $x + y = 5$, so $y_R = 5 - x_R$.Substituting $y_R$ into the centroid equation: $y_Q + (5 - x_R) = 9 \implies y_Q = 4 + x_R$.Also, $x_Q = 6 - x_R$.Substituting $x_Q$ and $y_Q$ into the circle equation: $(6 - x_R - 7)^2 + (4 + x_R - 7)^2 = 16$.$(-1 - x_R)^2 + (x_R - 3)^2 = 16$.$1 + x_R^2 + 2x_R + x_R^2 - 6x_R + 9 = 16$.$2x_R^2 - 4x_R - 6 = 0 \implies x_R^2 - 2x_R - 3 = 0$.Solving for $x_R$, we get $(x_R - 3)(x_R + 1) = 0$, so $x_R = 3$ or $x_R = -1$.The corresponding ordinates $y_R = 5 - x_R$ are $y_R = 5 - 3 = 2$ and $y_R = 5 - (-1) = 6$.The sum of the ordinates is $2 + 6 = 8$.

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