The equation of the hyperbola whose asymptotes are straight lines $3 x - 4 y + 7 = 0$ and $4 x + 3 y + 1 = 0$ and which passes through the origin is :

12 x^{2} - 7 x y - 12 y^{2} + 31 x + 17 y = 0

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12 x^{2} - 7 x y - 12 y^{2} + 31 x = 0

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2 x^{2} - 7 x y - 2 y^{2} + 31 x + 17 y = 0

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None of these

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Solution

The correct option is A $12 x^{2} - 7 x y - 12 y^{2} + 31 x + 17 y = 0$
The combined equation of the asmptotes is $(3 x - 4 y + 7) (4 x + 3 y + 1) = 0$
So, the combined equation of the hyperbola is $(3 x - 4 y + 7) (4 x + 3 y + 1) + λ = 0$ ...(1)
It passes through the origin,
$∴ 7 \times 1 + λ = 0$
$\Rightarrow λ = - 7$
Putting the value of $λ$ in (1), we get
$(3 x - 4 y + 7) (4 x + 3 y + 1) - 7 = 0$
$\Rightarrow 12 x^{2} - 7 x y - 12 y^{2} + 31 x + 17 y = 0$
This is the equation of the required hyperbola.

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12 x^{2} - 7 x y - 12 y^{2} = 0; 71 x^{2} + 94 x y - 71 y^{2} = 0; x^{2} - y^{2} = 0

The equation of the hyperbola whose asymptotes are the lines 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0 and which passes through origin is

3 x - 4 y + 7 = 0

4 x + 3 y + 1 = 0

(3 x - 4 y + 7) (4 x + 3 y + 1) = 7

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