Let lambda x - 2y = mu be a tangent to the hy | vedclass

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(D) The given hyperbola is $a^{2}x^{2} - y^{2} = b^{2}$,which can be written as $\frac{x^{2}}{(b/a)^{2}} - \frac{y^{2}}{b^{2}} = 1$.For a line $y = mx

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$4$

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(D) The given hyperbola is $a^{2}x^{2} - y^{2} = b^{2}$,which can be written as $\frac{x^{2}}{(b/a)^{2}} - \frac{y^{2}}{b^{2}} = 1$.For a line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$,the condition is $c^{2} = A^{2}m^{2} - B^{2}$.Here,the line is $\lambda x - 2y = \mu$,which can be rewritten as $y = \frac{\lambda}{2}x - \frac{\mu}{2}$.Comparing with $y = mx + c$,we have $m = \frac{\lambda}{2}$ and $c = -\frac{\mu}{2}$.Here $A^{2} = \frac{b^{2}}{a^{2}}$ and $B^{2} = b^{2}$.Substituting these into the condition $c^{2} = A^{2}m^{2} - B^{2}$:$(-\frac{\mu}{2})^{2} = \frac{b^{2}}{a^{2}}(\frac{\lambda}{2})^{2} - b^{2}$$\frac{\mu^{2}}{4} = \frac{b^{2}\lambda^{2}}{4a^{2}} - b^{2}$Multiply by $\frac{4}{b^{2}}$:$\frac{\mu^{2}}{b^{2}} = \frac{\lambda^{2}}{a^{2}} - 4$Rearranging the terms,we get $\frac{\lambda^{2}}{a^{2}} - \frac{\mu^{2}}{b^{2}} = 4$.

Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola $a^{2}x^{2} - y^{2} = b^{2}$. Then $\left(\frac{\lambda}{a}\right)^{2} - \left(\frac{\mu}{b}\right)^{2}$ is equal to

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