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If line $ax$ + $by$ = $1$ is normal to the hyperbola $\frac{{{x^2}}}{{{p^2}}} - \frac{{{y^2}}}{{{q^2}}} = 1$ then $\frac{{{p^2}}}{{{a^2}}} - \frac{{{q^2}}}{{{b^2}}} = 1$ is equal to (where $a$,$b$,$p$, $q \in {R^ + })$-
$0$
$1$
${\left( {{a^2} + {b^2}} \right)^2}$
${\left( {{p^2} + {q^2}} \right)^2}$
Solution
Let equation of normal
${\rm{P}}\cos \theta {\rm{x}} + {\rm{q}}\cot \theta {\rm{y}} = {{\rm{p}}^2} + {{\rm{q}}^2}$ ……$(1)$
given equaiton of normal $a x+b y=1$ ……$(2)$
$(1)$ and $( 2)$ are coincide
$\frac{p \cos \theta}{a}=\frac{q \cot \theta}{b}=p^{2}+q^{2}$
$\cos \theta=\frac{a}{p}\left(p^{2}+q^{2}\right)$ and $\sin \theta=\frac{a q}{b p}$
$\cos ^{2} \theta+\sin ^{2} \theta=1$
$\therefore \frac{\mathrm{p}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{q}^{2}}{\mathrm{b}^{2}}=\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)^{2}$
