The asymptotes of a hyperbola having center at the point (2, 3) are parallel to $3 x + 4 y = 0$ and $4 x + 3 y = 0$ If the hyperbola passes through (1, 2). then its equation is:

(3 x + 4 y - 18) (4 x + 3 y - 17) = 0

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(3 x + 4 y - 18) (4 x + 3 y - 17) = 49

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(3 x + 4 y - 18) (4 x + 3 y - 17) = 12

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(3 x + 4 y - 1) (4 x + 3 y - 2) = 49

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Solution

The correct option is B $(3 x + 4 y - 18) (4 x + 3 y - 17) = 49$
As the asymptotes are parallel to $3 x - 4 y = 0 a n d 4 x + 3 y = 0$ so their equations be given by $3 x - 4 y + λ_{1} = 0 a n d 4 x + 3 y + λ_{2} = 0$
Now asymptotes passes through the centre $(2, 3)$
$∴ 6 + 12 + λ_{1} a n d 8 + 8 + λ_{2} = 0$
$\Rightarrow λ_{1} = - 18 a n d λ_{2} = - 17$
so equation of asymptotes are $3 x + 4 y - 18 = 0$ and $4 x + 3 y - 17 = 0$
Now hyperbola and its ,asymptotes differ by the constant only, so the equation of hyperbola is given by
$(3 x + 4 y - 18) (4 x + 3 y - 17) + λ = 0$ which passes through the point $(1, 2)$ therefore $(3 (1) + 4 (2) - 18) (4 (1) + 3 (2) - 17) + λ = 0$
which gives $λ = - 49$
$∴$ The equation of hyperbola be
$(3 x + 4 y - 18) (4 x + 3 y - 17) = 49$
Hence, option 'B' is correct..

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