The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta+2 \cos…

The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r}$ is :
  1. $2=\sqrt{3} r \cos \theta-2 r \sin \theta$
  2. $5=-2 \sqrt{3} r \sin \theta+4 r \cos \theta$
  3. $2=\sqrt{3} r \cos \theta+2 r \sin \theta$
  4. $5=2 \sqrt{3} r \sin \theta+4 r \cos \theta$

Solution

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 $\sqrt{3} \cos \theta-2 \sin \theta=\frac{2}{r}$ $\therefore \quad \sqrt{3} r \cos \theta-2 r \sin \theta=2$

Asked in: AP EAMCET 2003

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