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

Question

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

Accepted Answer

Centroid :

Let $Q$ be a point on the given circle. First write the circle in standard form:

${x}^{2}+{y}^{2}-14x-14y+82=0\Rightarrow (x-7{)}^{2}+(y-7{)}^{2}=16$

So we can take

$Q(7+4\cos 	heta ,\, 7+4\sin 	heta )$

Point $R$ lies on the line $x+y=5$. Let its ordinate be $y$, then its abscissa is $5-y$, so

$R(5-y,\, y)$

Now use the centroid formula. For triangle $PQR$, centroid is

$(\frac{{x}_{P}+{x}_{Q}+{x}_{R}}{3},\, \frac{{y}_{P}+{y}_{Q}+{y}_{R}}{3})$

Given $P(3\cos \alpha ,\, 2\sin \alpha )$, so the centroid becomes

$\begin{gathered} (\frac{3\cos \alpha +7+4\cos 	heta +5-y}{3},\frac{2\sin \alpha +7+4\sin 	heta +y}{3}) \ =(\cos \alpha +2,\frac{2}{3}\sin \alpha +3) \end{gathered}$

On comparison

Compare the $x$-coordinates:

$\frac{3\cos \alpha +7+4\cos 	heta +5-y}{3}=\cos \alpha +2\Rightarrow 4\cos 	heta =y-6$

So,

$\cos 	heta =\frac{y-6}{4}$

Compare the $y$-coordinates:

$\frac{2\sin \alpha +7+4\sin 	heta +y}{3}=\frac{2}{3}\sin \alpha +3\Rightarrow 4\sin 	heta =2-y$

So,

$\sin 	heta =\frac{2-y}{4}$

Now use the identity ${\sin }^{2}	heta +{\cos }^{2}	heta =1$:

${\sin }^{2}	heta +{\cos }^{2}	heta =1\Rightarrow (y-6{)}^{2}+(y-2{)}^{2}=16$

Simplify:

${y}^{2}-8y+12=0$

This quadratic gives two possible values of $y$ (ordinates of $R$). The sum of roots is $8$, so

$\Rightarrow =8$

Solutions · 1