Let C be the circle x^2+(y-1)^2=2, and E_1,E_2 be two ellipses whose centres lie at the or…

Question

Let $C$ be the circle $x^2+(y-1)^2=2$, and $E_1,E_2$ be two ellipses whose centres lie at the origin with major axes along the $x$-axis and $y$-axis respectively. Let the line $x+y=3$ touch $C,E_1,E_2$ at $P,Q,R$ respectively. Given that $P$ is the midpoint of the segment $QR$, and $PQ=\dfrac{2\sqrt2}{3}$, the value of $9(x_1y_1+x_2y_2+x_3y_3)$ (for $P(x_1,y_1),Q(x_2,y_2),R(x_3,y_3)$) is equal to \_\_\_\_\_\_\_\_.

Answer: 46

Accepted Answer

Solving $x+y=3$ with $C$: $x^2+(2-x)^2=2 \Rightarrow x^2-2x+1=0 \Rightarrow x=1,\ y=2$. So $P=(1,2)$.

Since $P$ is the midpoint of $QR$: $x_2+x_3=2,\ y_2+y_3=4$. Let $Q=(a,b)$, so $R=(2-a,4-b)$.

$$PQ^2=(a-1)^2+(b-2)^2=\frac89 \;\Rightarrow\; 9a^2+9b^2-18a-36b+37=0,$$

which (with the tangency geometry) gives $a=\dfrac53,\ b=\dfrac43$.

$$x_1y_1+x_2y_2+x_3y_3 = 2+ab+(2-a)(4-b) = 2+\frac{20}{9}+\frac89=\frac{46}{9}.$$

$9(x_1y_1+x_2y_2+x_3y_3)=46.$

Solutions · 1