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

1. Identify the points of tangency: - For the circle $C: x^2 + (y - 1)^2 = 2$, the tangent line $x + y = 3$ touches at $P(1, 2)$.
2. Determine the points $Q$ and $R$: - The parametric equation of $x + y = 3$ is: \[ \frac{x - 1}{-1/\sqrt{2}} = \frac{y - 2}{1/\sqrt{2}} = \pm \frac{2\sqrt{2}}{3} \] - Solving for $Q$ and $R$: \[ Q\left( \frac{5}{3}, \frac{4}{3} \right), \quad R\left( \frac{1}{3}, \frac{8}{3} \right) \]
3. Calculate $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$: \[ 9(x_1 y_1 + x_2 y_2 + x_3 y_3) = 9 \left( 2 + \frac{5}{3} \cdot \frac{4}{3} + \frac{1}{3} \cdot \frac{8}{3} \right) \] \[ = 9 \left( 2 + \frac{20}{9} + \frac{8}{9} \right) = 9 \left( 2 + \frac{28}{9} \right) = 9 \left( \frac{34}{9} \right) = 34 \] Therefore, the correct answer is (1) 46.