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 the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to

Hint:

Use the parametric form of the tangent line to find the points of tangency.

Answer 1

Correct Answer: 46

We have the circle \(C: x^2 + (y-1)^2 = 2\), and the line \(x+y=3\) is tangent to \(C\) at \(P(x_1,y_1)\). The same line is tangent to two ellipses \(E_1\) (major axis along \(x\)-axis) and \(E_2\) (major axis along \(y\)-axis) at \(Q(x_2,y_2)\) and \(R(x_3,y_3)\) respectively. Given: \(P\) is the midpoint of \(QR\) and \(PQ=\dfrac{2\sqrt{2}}{3}\). We are to find \(9(x_1y_1+x_2y_2+x_3y_3)\).

Concept Used:

For a line \(ax+by+c=0\), the foot of the perpendicular from a point \((x_0,y_0)\) on the line is found by projection. If a line is tangent to a circle, the radius to the point of tangency is perpendicular to the tangent; hence the tangency point is the foot of the perpendicular from the center to the line. If a point \(P\) is the midpoint of segment \(QR\) on a fixed line, then \(Q\) and \(R\) are symmetric about \(P\) along that line; their coordinates are \(P\pm s\mathbf{u}\) where \(\mathbf{u}\) is a unit direction vector of the line and \(s\) is half the length of \(QR\). Here, \(PQ=s\).

Step-by-Step Solution:

Step 1: Find the tangency point \(P\) of line \(x+y=3\) with circle centered at \((0,1)\) of radius \(\sqrt{2}\).

\[ \text{Line: }x+y-3=0,\quad \text{center }C_0=(0,1),\quad n=(1,1). \] The foot of the perpendicular from \(C_0\) to the line is \[ P=C_0-\frac{n\cdot C_0-3}{\|n\|^2}\,n =(0,1)-\frac{1-3}{2}(1,1)=(0,1)+1\cdot(1,1)=(1,2). \] Thus \(P=(1,2)\Rightarrow x_1y_1=1\cdot 2=2.\)

 

Step 2: Parameterize points on the line \(x+y=3\) about \(P\), using a unit direction vector.

\[ \text{Direction vector }d=(1,-1),\quad \mathbf{u}=\frac{1}{\sqrt{2}}(1,-1). \] Given \(PQ=\dfrac{2\sqrt{2}}{3}\), write \[ Q=P+s\mathbf{u},\quad R=P-s\mathbf{u},\quad s=PQ=\frac{2\sqrt{2}}{3}. \] Compute the displacement: \[ s\mathbf{u}=\frac{2\sqrt{2}}{3}\cdot\frac{1}{\sqrt{2}}(1,-1)=\left(\frac{2}{3},-\frac{2}{3}\right). \] Hence \[ Q=\left(1+\frac{2}{3},\,2-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{4}{3}\right),\quad R=\left(1-\frac{2}{3},\,2+\frac{2}{3}\right)=\left(\frac{1}{3},\frac{8}{3}\right). \] Therefore \[ x_2y_2=\frac{5}{3}\cdot\frac{4}{3}=\frac{20}{9},\qquad x_3y_3=\frac{1}{3}\cdot\frac{8}{3}=\frac{8}{9}. \]

 

Step 3: Sum and scale.

\[ x_1y_1+x_2y_2+x_3y_3 =2+\frac{20}{9}+\frac{8}{9} =2+\frac{28}{9} =\frac{46}{9}. \] \[ 9(x_1y_1+x_2y_2+x_3y_3)=9\cdot\frac{46}{9}=46. \]

 

Final Computation & Result

The required value is \(46\).