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The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r}$ is :
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Verified Answer
The correct answer is:
$2=\sqrt{3} r \cos \theta-2 r \sin \theta$
Given,
$\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r}$ ...(i)
Any line perpendicular to Eq. (i) is
$\sqrt{3} \sin \left(\frac{\pi}{2}+\theta\right)+2 \cos \left(\frac{\pi}{2}+\theta\right)=\frac{k}{r}$
$\sqrt{3} \cos \theta-2 \sin \theta=\frac{k}{r}$
It passes through $\left(-1, \frac{\pi}{2}\right)$, then
$\sqrt{3} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{2}=\frac{k}{-1}$
$-2=\frac{k}{-1} \Rightarrow k=2$
Thus, the equation is
$\sqrt{3} \cos \theta-2 \sin \theta=\frac{2}{r}$
$\therefore \quad \sqrt{3} r \cos \theta-2 r \sin \theta=2$
$\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r}$ ...(i)
Any line perpendicular to Eq. (i) is
$\sqrt{3} \sin \left(\frac{\pi}{2}+\theta\right)+2 \cos \left(\frac{\pi}{2}+\theta\right)=\frac{k}{r}$
$\sqrt{3} \cos \theta-2 \sin \theta=\frac{k}{r}$
It passes through $\left(-1, \frac{\pi}{2}\right)$, then
$\sqrt{3} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{2}=\frac{k}{-1}$
$-2=\frac{k}{-1} \Rightarrow k=2$
Thus, the equation is
$\sqrt{3} \cos \theta-2 \sin \theta=\frac{2}{r}$
$\therefore \quad \sqrt{3} r \cos \theta-2 r \sin \theta=2$
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