The equation of the hyperbola with asymptotes | vedclass

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Question

(C) The joint equation of the asymptotes is given by $(3x - 4y + 7)(4x + 3y + 1) = 0$. The equation of the hyperbola differs from the joint equation o

Vedclass · Accepted Answer

$12x^2 - 7xy - 12y^2 + 31x + 17y = 0$

Vedclass · Answer

(C) The joint equation of the asymptotes is given by $(3x - 4y + 7)(4x + 3y + 1) = 0$. The equation of the hyperbola differs from the joint equation of its asymptotes by a constant $k$. Thus,the equation of the hyperbola is $(3x - 4y + 7)(4x + 3y + 1) + k = 0$. Since the hyperbola passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation: $(3(0) - 4(0) + 7)(4(0) + 3(0) + 1) + k = 0 (7)(1) + k = 0 k = -7$. Substituting $k = -7$ back into the equation: $(3x - 4y + 7)(4x + 3y + 1) - 7 = 0 (12x^2 + 9xy + 3x - 16xy - 12y^2 - 4y + 28x + 21y + 7) - 7 = 0 12x^2 - 7xy - 12y^2 + 31x + 17y = 0$.

The equation of the hyperbola with asymptotes $3x - 4y + 7 = 0$ and $4x + 3y + 1 = 0$ and which passes through the origin is

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