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

Question

Let $C$ be the circle $x^2+(y-1)^2=2, E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on x -axis and y -axis respectively. Let the straight line $x+y=3$ touch the curves $C, E_1$ and $E_2$

Accepted Answer

Solving the line $x+y=3$, and the circle ${x}^{2}+$ $(y-1{)}^{2}=2$

Substitute $y=3-x$ :

$\begin{gathered} {x}^{2}+(3-x-1{)}^{2}=2 \ \Rightarrow {x}^{2}-2x+1=0 \ \Rightarrow x=1\Rightarrow y=2 \end{gathered}$

So, $P=({x}_{1},{y}_{1})=(1,2)\Rightarrow {x}_{1}{y}_{1}=1\cdot 2=2$

Use midpoint condition

Let $Q=({x}_{2},{y}_{2}),R=({x}_{3},{y}_{3})$.

Since $P$ is the midpoint of QR:

$x_2+x_3=2 x_1=2, y_2+y_3=2 y_1=4$So, we can write: ${x}_{3}=2-{x}_{2},{y}_{3}=4-{y}_{2}$

Given,

$PQ=\frac{2\sqrt{2}}{3}\Rightarrow P{Q}^{2}={({x}_{2}-1)}^{2}+{({y}_{2}-2)}^{2}=\frac{8}{9}$

Let's denote: ${x}_{2}=a,{y}_{2}=b,{x}_{3}=2-a,{y}_{3}=4-b$

$\begin{gathered} (a-1{)}^{2}+(b-2{)}^{2}=\frac{8}{9} \ \Rightarrow {a}^{2}-2a+1+{b}^{2}-4b+4=\frac{8}{9} \ \Rightarrow {a}^{2}+{b}^{2}-2a-4b+5=\frac{8}{9} \ \Rightarrow 9{a}^{2}+9{b}^{2}-18a-36b+37=0 \end{gathered}$

Hence, $a=\frac{5}{3},b=\frac{4}{3}$

$\begin{gathered} {x}_{1}{y}_{1}+{x}_{2}{y}_{2}+{x}_{3}{y}_{3}=2+ab+(2-a)(4-b) \ 9({x}_{1}{y}_{1}+{x}_{2}{y}_{2}+{x}_{3}{y}_{3})=9(10+2ab-2b-4a) \ =90+18ab-18b-36a=46 \end{gathered}$

Solutions · 1