If a hyperbola has asymptotes 3x - 4y - 1 = 0 | vedclass

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(A) The equation of a hyperbola with asymptotes $L_1 = 0$ and $L_2 = 0$ is given by $L_1 L_2 = k$,where $k$ is a constant. Given asymptotes are $L_1:

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$x + y - 5 = 0, x - y - 1 = 0$

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(A) The equation of a hyperbola with asymptotes $L_1 = 0$ and $L_2 = 0$ is given by $L_1 L_2 = k$,where $k$ is a constant. Given asymptotes are $L_1: 3x - 4y - 1 = 0$ and $L_2: 4x - 3y - 6 = 0$. The axes of the hyperbola are the angle bisectors of the asymptotes. The equations of the angle bisectors are given by $\frac{3x - 4y - 1}{\sqrt{3^2 + (-4)^2}} = \pm \frac{4x - 3y - 6}{\sqrt{4^2 + (-3)^2}}$. Since the denominators are equal,we have $3x - 4y - 1 = \pm (4x - 3y - 6)$. Case $1$: $3x - 4y - 1 = 4x - 3y - 6 \implies x + y - 5 = 0$. Case $2$: $3x - 4y - 1 = -(4x - 3y - 6) \implies 3x - 4y - 1 = -4x + 3y + 6 \implies 7x - 7y - 7 = 0 \implies x - y - 1 = 0$. Thus,the axes are $x + y - 5 = 0$ and $x - y - 1 = 0$.

If a hyperbola has asymptotes $3x - 4y - 1 = 0$ and $4x - 3y - 6 = 0$,then the transverse and conjugate axes of that hyperbola are

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