Question

Let $A(\alpha ,\beta ),B({\alpha }^2,\beta ),C(\alpha ,{\beta }^2)$ are three ditinct points which are at same distance from origin. Then the sum of all possible value of ${}^{\prime }{\theta }^{\prime }$ such that $(\sin \theta ,\cos \theta )$ is equidistant to any of these points taken pairwise is
where $(0\le \theta \le \frac{\pi }{2})$

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

The correct option is B $\frac{3\pi }{4}$

Since points $A,B,C$ are at the same distance from origin
$\Rightarrow {\alpha }^2+{\beta }^2={\alpha }^4+{\beta }^2={\alpha }^2+{\beta }^4$, Possible values for $\alpha ,\beta$ : $0,1,-1$
with ${}^{\prime }{0}^{\prime }$ including distinct points are not possible
$\therefore A(-1,-1),B(1,-1)C(-1,1)$ is the only possible case here
Let $P(\sin \theta ,\cos \theta )$
$PA=PB\Rightarrow (\sin \theta +1{)}^2+(\cos \theta +1{)}^2=(\sin \theta -1{)}^2+(\cos \theta +1{)}^2$
$\Rightarrow \sin \theta =0\Rightarrow \theta =0^{\circ }$
$BP=PC\Rightarrow (\sin \theta -1{)}^2+(\cos \theta +1{)}^2=(\sin \theta +1{)}^2+(\cos \theta -1{)}^2$
$\Rightarrow \sin \theta =\cos \theta \Rightarrow \theta =\frac{\pi }{4}$
$AP=PC\Rightarrow (\sin \theta +1{)}^2+(\cos \theta +1{)}^2=(\sin \theta +1{)}^2+(\cos \theta -1{)}^2$
$\Rightarrow \cos \theta =0$
$\Rightarrow \theta =\frac{\pi }{2}$

Hence required sum is $=\frac{3\pi }{4}$