If lx + my = 1 is a normal to the hyperbola f | vedclass

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(D) The given equation of the normal is $lx + my = 1$,which can be rewritten as $y = -(\frac{l}{m})x + \frac{1}{m}$.Comparing this with the slope-inte

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$l^2 m^2(a^2 + b^2)^2$

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(D) The given equation of the normal is $lx + my = 1$,which can be rewritten as $y = -(\frac{l}{m})x + \frac{1}{m}$.Comparing this with the slope-intercept form $y = Mx + C$,we have $M = -\frac{l}{m}$ and $C = \frac{1}{m}$.The condition for the line $y = Mx + C$ to be a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $C^2 = \frac{M^2(a^2 + b^2)^2}{a^2 M^2 - b^2}$.Substituting the values of $M$ and $C$:$(\frac{1}{m})^2 = \frac{(-\frac{l}{m})^2(a^2 + b^2)^2}{a^2(-\frac{l}{m})^2 - b^2}$$\frac{1}{m^2} = \frac{\frac{l^2}{m^2}(a^2 + b^2)^2}{\frac{a^2 l^2 - b^2 m^2}{m^2}}$$\frac{1}{m^2} = \frac{l^2(a^2 + b^2)^2}{a^2 l^2 - b^2 m^2}$$a^2 l^2 - b^2 m^2 = l^2 m^2(a^2 + b^2)^2$.

If $lx + my = 1$ is a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then $a^2 m^2 - b^2 l^2 =$

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