The equation of a hyperbola whose asymptotes | vedclass

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(D) The equation of the asymptotes of the hyperbola is given by $3x \pm 5y = 0$.The equation of the hyperbola can be written as $(3x + 5y)(3x - 5y) =

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$9x^2 - 25y^2 = 225$

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(D) The equation of the asymptotes of the hyperbola is given by $3x \pm 5y = 0$.The equation of the hyperbola can be written as $(3x + 5y)(3x - 5y) = \lambda$,where $\lambda$ is a constant.This simplifies to $9x^2 - 25y^2 = \lambda$.Since the vertices of the hyperbola are $(\pm 5, 0)$,the point $(5, 0)$ must satisfy the equation of the hyperbola.Substituting $(x, y) = (5, 0)$ into the equation $9x^2 - 25y^2 = \lambda$,we get:$9(5)^2 - 25(0)^2 = \lambda$$9(25) = \lambda$$\lambda = 225$Therefore,the equation of the hyperbola is $9x^2 - 25y^2 = 225$.

The equation of a hyperbola whose asymptotes are $3x \pm 5y = 0$ and vertices are $(\pm 5, 0)$ is

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