If the eccentricity of the hyperbola frac{x^2 | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. It is a standard property that the area of the triangle formed by the asympt

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$ab$

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(B) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. It is a standard property that the area of the triangle formed by the asymptotes of a hyperbola with any tangent is constant and equal to $ab$. Given the eccentricity $e = \sec \alpha$,we have $e^2 = 1 + \frac{b^2}{a^2} = \sec^2 \alpha$. Since $1 + 	an^2 \alpha = \sec^2 \alpha$,we get $\frac{b^2}{a^2} = 	an^2 \alpha$,which implies $b = a 	an \alpha$. Substituting this into the area formula,the area is $a(a 	an \alpha) = a^2 	an \alpha$. However,the standard result for the area is $ab$. Given $b = a 	an \alpha$,the area is $a^2 	an \alpha$. Comparing with the options,if we evaluate $ab$,we get $a(a 	an \alpha) = a^2 	an \alpha$. Since the question asks for the area in terms of the given parameters,and $ab$ is the constant area,the correct value is $ab$.

If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sec \alpha$,then the area of the triangle formed by the asymptotes of the hyperbola with any of its tangent is

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