Let P(3, 2), 0, be a point on the ellipse x^29+y^24=1; let Q be a point on the circle x^2+…

Question

Let $P(3\cos\alpha, 2\sin\alpha),\ \alpha
e0$, be a point on the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$; let $Q$ be a point on the circle $x^2+y^2-14x-14y+82=0$; and let $R$ be a point on the line $x+y=5$ such that the centroid of triangle $PQR$ is $\left(\cos\alpha+2,\ \dfrac23\sin\alpha+3ight)$. Then the sum of the ordinates of all possible points $R$ is:

Accepted Answer

The circle is $(x-7)^2+(y-7)^2=16$, so $Q=(7+4\cos\theta,\,7+4\sin\theta)$. On the line $x+y=5$, write $R=(5-y,\,y)$.

Centroid $x$-coordinate: $\dfrac{3\cos\alpha+7+4\cos\theta+5-y}{3}=\cos\alpha+2 \Rightarrow 4\cos\theta=y-6$.

Centroid $y$-coordinate: $\dfrac{2\sin\alpha+7+4\sin\theta+y}{3}=\dfrac23\sin\alpha+3 \Rightarrow 4\sin\theta=2-y$.

Using $\sin^2\theta+\cos^2\theta=1$:

$$(y-6)^2+(y-2)^2=16 \;\Rightarrow\; y^2-8y+12=0.$$

Sum of the two roots (ordinates of $R$) $=8$.

Solutions · 1