If the distance between the points $(a \cos \theta, a \sin \theta)$ and $(a \cos \phi, a \sin \phi)$ is $2…
If the distance between the points $(a \cos \theta, a \sin \theta)$ and $(a \cos \phi, a \sin \phi)$ is $2 a$, then $\theta$ is equal to
- $2 n \pi \pm \pi+\phi, n \in Z$
- $n \pi+\frac{\pi}{2}+\phi, n \in Z$
- $n \pi-\phi, n \in Z$
- $2 n \pi+\phi, n \in Z$
Solution
Let the points be $A=(a \cos \theta, a \sin \theta)$ and
$
\begin{aligned}
& B=(a \cos \phi, a \sin \phi) \\
& \therefore \quad A B=\sqrt{(a \cos \theta-a \cos \phi)^2}+(a \sin \theta-a \sin \phi)^2 \\
& =\sqrt{a^2 \cos ^2 \theta+a^2 \cos ^2 \phi-2 a^2 \cos \theta \cos \phi} \\
& \quad \quad+a^2 \sin ^2 \theta+a^2 \sin ^2 \phi-2 a^2 \sin \theta \sin \phi
\end{aligned}
$
$
\begin{aligned}
& =\sqrt{2 a^2-2 a^2(\cos \theta \cos \phi+\sin \theta \sin \phi)} \\
& =\sqrt{2} a(\sqrt{1-\cos (\theta-\phi)} \\
& \Rightarrow \quad 2 a=\sqrt{2} a \sqrt{2} \sin \left(\frac{\theta-\phi}{2}\right) \\
& \Rightarrow \quad \sin \left(\frac{\theta-\phi}{2}\right)=1 \Rightarrow \frac{\theta-\phi}{2}=n \pi \pm \frac{\pi}{2} \\
& \Rightarrow \quad \sin \left(\frac{\theta-\phi}{2}\right)=1 \Rightarrow \frac{\theta-\phi}{2}=n \pi \pm \frac{\pi}{2} \\
& \Rightarrow \quad \theta-\phi=2 n \pi \pm \pi \\
& \Rightarrow \quad \theta=2 n \pi \pm \pi+\phi \\
&
\end{aligned}
$
where, $n \in Z$
Asked in: AP EAMCET 2004
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