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Hence, the joint equation of the asymptotes is (3x – 4y + 7)(4x + 3y + 1) = 0.
Now, we know that the equation of rectangular hyperbola differs from that of the joint equation of asymptotes by a constant only, hence we must have the equation of the hyperbola as
(3x – 4y + 7)(4x + 3y + 1) + r = 0.
It is given that the hyperbola passes through the origin and hence, putting x = y = 0 , we obtain,
7+ r = 0
Hence, r = -7.
Hence, the equation of the hyperbola is (3x – 4y + 7)(4x + 3y + 1) – 7= 0.
This gives 12x2 + 9xy + 3x -16xy – 12y2 – 4y + 28x + 21y + 7- 7 = 0.
So, 12x2 – 12y2 - 7xy + 31x + 17y = 0
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