Question

A circle whose center coincides with the origin having radius ${}^{\prime }{a}^{\prime }$ cuts x-axis at $A$ and $B$. If $P$ and $Q$ are two points on the circle whose parametric angles differ by $2\theta$, then locus of the intersection pont $AP$ and $BQ$ is

• A. $x^2+y^2+2ay\tan \theta =a^2$
• B. $x^2+y^2-2ay\tan \theta =a^2$
• C. $x^2+y^2+2ay\cot \theta =a^2$
• D. None of these

Answer 1

Answer: (A)

$x^2+y^2+2ay\tan \theta =a^2$

Let $P\equiv \left(a\cos \alpha ,a\sin \alpha \right)$ and $Q\equiv (a\cos \beta ,a\sin \beta ),$ where $\beta -\alpha =2\theta$

Also, $A\equiv (a,0)$ and $B\equiv (-a,0)$

If $R(h,k)$ be the intersection point of $AP$ and $BQ,$

the slope of $AR=$ slope of $AP$ $[\because R$ is lies on $AP]$

$\Rightarrow \frac{k}{h-a}=\frac{\sin \alpha }{\cos \alpha -1}\Rightarrow \tan \left(\frac{\alpha }{2}\right)=\frac{a-h}{h}$ ...(1)

$\Rightarrow \frac{k}{h+a}=\frac{\sin \beta }{\cos \beta +1}\Rightarrow \tan \left(\frac{\beta }{2}\right)=\frac{k}{h+a}$ ...(2)

Since, $\beta -\alpha =2\theta ,$ we have $\frac{\beta }{2}-\frac{\alpha }{2}=\theta$

$\Rightarrow \frac{\tan \left(\frac{\beta }{2}\right)-\tan \left(\frac{\alpha }{2}\right)}{1+\tan \left(\frac{\beta }{2}\right)\tan \left(\frac{\alpha }{2}\right)}=\tan \theta$

$\Rightarrow \frac{\frac{k}{h+a}-\frac{a-h}{k}}{1+\left(\frac{k}{h+a}\right)\left(\frac{a-h}{k}\right)}=\tan \theta$

$\Rightarrow h^2+k^2-2ak\tan \theta =a^2$

Hence, the locus of $R$ is $x^2+y^2-2ay\tan \theta =a^2.$