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If the normal form of the equation of a straight line $4 x+3 y+2=0$ is $x \cos \alpha+y \sin \alpha=p$ and its intercept form is $\frac{x}{a}+\frac{y}{b}=1$, then $\frac{p \sec \alpha}{a b}=$
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$\frac{-3}{2}$
$\begin{gathered}\text { Given, } 4 x+3 y+2=0 \Rightarrow-4 x-3 y=2 \\ \frac{-4 x}{5}-\frac{3}{5} y=\frac{2}{5} \\ \because \cos \alpha=\frac{-4}{5}, \sin \alpha=\frac{-3}{5} \text { and } P=\frac{2}{5} \\ \frac{x}{2 /-4}+\frac{y}{2 /-3}=1 \Rightarrow a=\frac{-2}{4}, b=\frac{-2}{3} \\ \therefore \quad \frac{P \sec \alpha}{a b}=\frac{\frac{2}{5} \times\left(\frac{-5}{4}\right)}{\left(\frac{-2}{4}\right)\left(\frac{-2}{3}\right)}=\frac{-3}{2}\end{gathered}$
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