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Question

# The equation of the hyperbola whose asymptotes are the straight lines 3x â€“ 4y + 7 = 0 and 4x + 3y + 1 = 0 and which passes through the origin, is

A
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B
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C
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D
None of the above
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Solution

## The correct option is A The combined equation of the asymptotes is (3x -4y + 7)(4x + 3y + 1) = 0 So, the combined equation of the hyperbola is (3x−4y+7)(4x+3y+1)+λ=0 It passes through the origin. ∴ 7×1+λ=0⇒λ=−7 On putting the value of \lambda in equation (i), we get (3x - 4y + 7)(4x + 3y + 1) - 7 = 0 ⇒12x2−7xy−12y2+31x+17y=0, which is the equation of the required hyperbola

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