Find the equation of the hyperbola which has `3x-4y+7=0` and `4x+3y+1=0` as its asymptotes and which passes through the origin.

Options

A. `12x^(2)-7xy-12y^(2)+31x+17y=0`
B. `12x^(2)-7xy+12y^(2)+31x+17y=0`
C. `12x^(2)+7xy-12y^(2)-31x+17y=0`
D. `12x^(2)-7xy-12y^(2)+31x-17y=0`

Explanation

The combined equation of the asymptotes is
`(3x-4y+7)(4x+3y+1)`
`=12x^(2)-7xy-12y^(2)+31x+17y+7=0 " " ....(1)`
Since the equation of the hyperbola and the combined equation of its asymptotes differ by merely a constant, the equation of the hyperbola may be taken as
`12x^(2)-7xy-12y^(2)+31x+17y +k = 0 " " ....(2)`
AS (ii) passes through (0,0), we have k = 0 . Hence, the equation to the required hyperbola is
`12x^(2)-7xy-12y^(2)+31x+17y=0`.