Let the eccentricity of the hyperbola H : fra | vedclass

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(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given $m

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$20$

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(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given $m = 2$,we have $c^2 = 4a^2 - b^2$.The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2}$.Substituting $e^2 = \frac{5}{2}$,we get $\frac{5}{2} = 1 + \frac{b^2}{a^2}$,which implies $\frac{b^2}{a^2} = \frac{3}{2}$,so $b^2 = \frac{3a^2}{2}$.The length of the latus rectum is $\frac{2b^2}{a} = 6\sqrt{2}$.Substituting $b^2 = \frac{3a^2}{2}$,we get $\frac{2}{a} 	imes \frac{3a^2}{2} = 6\sqrt{2}$,which simplifies to $3a = 6\sqrt{2}$,so $a = 2\sqrt{2}$.Then $a^2 = 8$ and $b^2 = \frac{3}{2} 	imes 8 = 12$.Finally,$c^2 = 4a^2 - b^2 = 4(8) - 12 = 32 - 12 = 20$.

Let the eccentricity of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and the length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$,then the value of $c^2$ is equal to

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