Asymptotes of the hyperbola $x y = 4 x + 3 y$ are

x=3, y=4

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x=4, y=3

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x=2, y=6

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x=6, y=2

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Solution

The correct option is A x=3, y=4
correct answer is A.
Given,

$x y = 4 x + 3 y$
$\Rightarrow (x - y) = 3 (y - 4) + 12$
$\Rightarrow (x - 4) (y - 4) = 12$
$∴$ joint equation of asymptotes is $(x - 3) (y - 4) = 0$
hence,
$x - 3 = 0$
$x = 3$
and,
$y - 4 = 0$

$y = 4$

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