Let x- y be a pair of real number satisfying 56 x -33 y - y - x 2- y 2 and33 x -56 y - x - x 2- y 2 - If - x - y - q P where p and q are relatively prime- then 6 p - q is

56x +33y = y/x^2+y^2 ⋯ (1) 33x – 56y = x/x^2+y^2 ⋯ (2) Multiply equation (1) by i and add to equation (2)= 1/33 + 56i⇒ z = ±1/7+4i=±(7 4i)/65 ∴ |x| + ...

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Standard XII

Mathematics

Directrix

Let x, y be a...

Question

Let (x, y) be a pair of real number satisfying
$56 x + 33 y = \frac{- y}{x^{2} + y^{2}}$ and
$33 x - 56 y = \frac{x}{x^{2} + y^{2}}$ . If |x| + |y| = $\frac{p}{q}$ (where p and q are relatively prime), then (6p – q)is ___

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Solution

$56 x + 33 y = \frac{- y}{x^{2} + y^{2}} \dots$ (1)
$33 x - 56 y = \frac{x}{x^{2} + y^{2}} \dots$ (2)
Multiply equation (1) by i and add to equation
$(2) = \frac{1}{33 + 56 i} \Rightarrow z = \pm \frac{1}{7 + 4 i} = \pm \frac{(7 - 4 i)}{65} ∴ | x | + | y | = \frac{11}{65}$

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Let (x, y) be a pair of real number satisfying 56x+33y=−yx2+y2 and 33x–56y=xx2+y2. If |x| + |y| = pq (where p and q are relatively prime), then (6p – q)is ___

56x+33y=−yx2+y2 ⋯ (1) 33x–56y=xx2+y2 ⋯ (2) Multiply equation (1) by i and add to equation (2)=133+56i⇒z=±17+4i=±(7−4i)65∴ |x|+|y|=1165

Let (x, y) be a pair of real number satisfying
$56 x + 33 y = \frac{- y}{x^{2} + y^{2}}$ and
$33 x - 56 y = \frac{x}{x^{2} + y^{2}}$ . If |x| + |y| = $\frac{p}{q}$ (where p and q are relatively prime), then (6p – q)is ___

$56 x + 33 y = \frac{- y}{x^{2} + y^{2}} \dots$ (1)
$33 x - 56 y = \frac{x}{x^{2} + y^{2}} \dots$ (2)
Multiply equation (1) by i and add to equation
$(2) = \frac{1}{33 + 56 i} \Rightarrow z = \pm \frac{1}{7 + 4 i} = \pm \frac{(7 - 4 i)}{65} ∴ | x | + | y | = \frac{11}{65}$