The asymptotes of the hyperbola frac{x^2}{a^2 | vedclass

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(A) The area of the triangle formed by the asymptotes and any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by the constan

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$\sec \lambda$

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(A) The area of the triangle formed by the asymptotes and any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by the constant value $ab$.Given that the area is $a^2 	an \lambda$,we have $ab = a^2 	an \lambda$.Dividing by $a^2$,we get $\frac{b}{a} = 	an \lambda$.The eccentricity $e$ of a hyperbola is given by $e^2 = 1 + \frac{b^2}{a^2}$.Substituting $\frac{b}{a} = 	an \lambda$,we get $e^2 = 1 + 	an^2 \lambda$.Using the identity $1 + 	an^2 \lambda = \sec^2 \lambda$,we have $e^2 = \sec^2 \lambda$.Therefore,$e = \sec \lambda$ (since $e > 1$ and $\sec \lambda$ is positive for the relevant range).

The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ form with any tangent to the hyperbola a triangle whose area is $a^2 \tan \lambda$ in magnitude. Then its eccentricity $e$ is:

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