If $A$ is the orthocenter of the triangle formed by $2 x^{2} - y^{2} = 0, x + y - 1 = 0$ , and $B$ is the centroid of the triangle formed by $2 x^{2} - 5 x y + 2 y^{2} = 0, 7 x - 2 y - 12 = 0,$ then the distance between $A$ and $B$ is

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$\sqrt{5}$

$\sqrt{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 2 x^{2} - y^{2} = 0 a n d x + y - 1 = 0 \\ a = 2, h = 0, b = - 1 a n d l = 1, m = 1, n = - 1 \\ K = \frac{- n (a + b)}{a m^{2} - 2 h l m + b l^{2}} = \frac{1 (2 - 1)}{2 (1) - 0 - 1} = 1 \\ H = o r t h o c e n t r e = (k l, k m) = (1, 1) \\ G i v e n e q u a t i o n 2 x^{2} - 5 x y + 2 y^{2} = 0 \\ a n d 7 x - 2 y - 12 = 0 \\ l e t f = 2 x^{2} + 5 x y + 2 y^{2}, l = 7, m = - 2, n = - 12 \end{array}$

$\begin{array}{l} \frac{\partial f}{\partial x} = 4 x + 5 y, \frac{\partial f}{\partial y} = 5 x + 4 y \\ f = {(2 x^{2} + 5 x y + 2 y^{2})}_{(l, m)} = 2 (49) - 70 + 8 \Rightarrow 98 - 70 + 8 \Rightarrow 36 \\ ∴ G = \frac{- n}{3 f} (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) a t x = l, y = m \\ = \frac{12}{3 (36)} (4 (7) + 5 (- 2), 5 (7) + 4 (- 2)) \Rightarrow \frac{1}{9} (18, 27) \Rightarrow (2, 3) \\ H G = \sqrt{{(2 - 1)}^{2} + {(3 - 1)}^{2}} = \sqrt{1 + 4} = \sqrt{5} \end{array}$