If the line ax + by = 1 is normal to the hype | vedclass

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(D) The equation of the normal to the hyperbola $\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$ at the point $(p \sec 	heta, q 	an 	heta)$ is given by $\fr

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$(p^2 + q^2)^2$

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(D) The equation of the normal to the hyperbola $\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$ at the point $(p \sec 	heta, q 	an 	heta)$ is given by $\frac{px}{\sec 	heta} + \frac{qy}{	an 	heta} = p^2 + q^2$,which simplifies to $px \cos 	heta + qy \cot 	heta = p^2 + q^2$.Comparing this with the given normal equation $ax + by = 1$,we can rewrite the given equation as $(p^2 + q^2)ax + (p^2 + q^2)by = p^2 + q^2$.Equating the coefficients,we get $p \cos 	heta = a(p^2 + q^2)$ and $q \cot 	heta = b(p^2 + q^2)$.Thus,$\cos 	heta = \frac{a(p^2 + q^2)}{p}$ and $\cot 	heta = \frac{b(p^2 + q^2)}{q}$.Using the identity $\csc^2 	heta - \cot^2 	heta = 1$,we have $\frac{1}{\cos^2 	heta} - \cot^2 	heta = 1$ is incorrect; rather,we use $\sec^2 	heta - 	an^2 	heta = 1$.From the expressions,$\sec 	heta = \frac{p}{a(p^2 + q^2)}$ and $	an 	heta = \frac{q}{b(p^2 + q^2)}$.Substituting into $\sec^2 	heta - 	an^2 	heta = 1$,we get $\frac{p^2}{a^2(p^2 + q^2)^2} - \frac{q^2}{b^2(p^2 + q^2)^2} = 1$.Therefore,$\frac{p^2}{a^2} - \frac{q^2}{b^2} = (p^2 + q^2)^2$.

If the line $ax + by = 1$ is normal to the hyperbola $\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$,then $\frac{p^2}{a^2} - \frac{q^2}{b^2}$ is equal to (where $a, b, p, q \in R^+$):

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