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The product of $\mathrm{y}$ the perpendiculars from the two points $(\pm 4,0)$ to the line $3 x \cos \phi+5 y \sin \phi=15$ is
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9
If length of perpendicular be $\mathrm{p}_{1}$ from the point $(4,0)$ $\begin{aligned} \mathrm{p}_{1} &=\left|\frac{12 \cos \phi-15}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}}\right| \\ &=\frac{15-12 \cos \phi}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}} \end{aligned}$
If length of perpendicular be $\mathrm{p}_{2}$ from the point $(-4,0)$
$p_{2}=\left|\frac{-12 \cos \phi-15}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}}\right|$
$\begin{aligned}=& \frac{(12 \cos \phi+15)}{\sqrt{(3 \cos \phi)^{2}+(5 \sin \phi)^{2}}} \\ \mathrm{p}_{1} \cdot \mathrm{p}_{2} &=\frac{(15-12 \cos \phi)(12 \cos \phi+15)}{(3 \cos \phi)^{2}+(5 \sin \phi)^{2}} \\ &=\frac{\left(225-144 \cos ^{2} \phi\right)}{9 \cos ^{2} \phi+25 \sin ^{2} \phi}=\frac{9\left(25-16 \cos ^{2} \phi\right)}{25-16 \cos ^{2} \phi} \\ &=9 \end{aligned}$
If length of perpendicular be $\mathrm{p}_{2}$ from the point $(-4,0)$
$p_{2}=\left|\frac{-12 \cos \phi-15}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}}\right|$
$\begin{aligned}=& \frac{(12 \cos \phi+15)}{\sqrt{(3 \cos \phi)^{2}+(5 \sin \phi)^{2}}} \\ \mathrm{p}_{1} \cdot \mathrm{p}_{2} &=\frac{(15-12 \cos \phi)(12 \cos \phi+15)}{(3 \cos \phi)^{2}+(5 \sin \phi)^{2}} \\ &=\frac{\left(225-144 \cos ^{2} \phi\right)}{9 \cos ^{2} \phi+25 \sin ^{2} \phi}=\frac{9\left(25-16 \cos ^{2} \phi\right)}{25-16 \cos ^{2} \phi} \\ &=9 \end{aligned}$
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