Let C be the circle , and 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, and at , and respectively. Given that P is the mid-point of the line

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 $E_{2}$ 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 $P Q = \frac{2 \sqrt{2}}{3}$ , the value of $9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})$ is equal to __________. [2025]

Smart Achievers · Accepted Answer

(46)

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

Substitute y = 3 – x

$x^{2} + {(3 - x - 1)}^{2} = 2$

$\Rightarrow x^{2} - 2 x + 1 = 0$

$\Rightarrow x = 1 \Rightarrow y = 2$

So, $P = (x_{1}, y_{1}) = (1, 2) \Rightarrow x_{1} y_{1} = 2$

Use mid-point condition

Let $Q = (x_{2}, y_{2})$ , $R = (x_{3}, y_{3})$ .

Since P is the mid-point of QR

$\Rightarrow x_{2} + x_{3} = 2 x_{1} = 2, y_{2} + y_{3} = 2 y_{1} = 4$

So, we can write : $x_{3} = 2 - x_{2}, y_{3} = 4 - y_{2}$

(b) Given

$P Q = \frac{2 \sqrt{2}}{3} \Rightarrow P Q^{2} = {(x_{2} - 1)}^{2} + {(y_{2} - 2)}^{2} = \frac{8}{9}$

Let's denote : $x_{2} = a, y_{2} = b, x_{3} = 2 - a, y_{3} = 5 - b$

${(a - 1)}^{2} + {(b - 2)}^{2} = \frac{8}{9}$

$\Rightarrow a^{2} - 2 a + 1 + b^{2} - 4 b + 4 = \frac{8}{9}$

$\Rightarrow a^{2} + b^{2} - 2 a - 4 b + 5 = \frac{8}{9}$

$\Rightarrow 9 a^{2} + 9 b^{2} - 18 a - 36 b + 37 = 0$

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

$x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} = 2 + a b + (2 - a) (4 - b)$

$9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3}) = 9 (10 + 2 a b - 2 b - 4 a)$

= 90 + 18ab – 18b – 36a = 46.

Solution

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