Question

For any value of $\theta$. if the straight lines $x\sin \theta +(1-\cos \theta )y=a\sin \theta$ and $x\sin \theta -(1+\cos \theta )y+a\sin \theta =0$ intersect at
$P(\theta )$. then the locus of $P(\theta )$ is a

• A. straight line
• B. circle
• C. parabola
• D. hyperbola

Answer 1

Answer: (B)

Given, equations of straight lines are< br>
$x\sin \theta +(1-\cos \theta )y=a\sin \theta ....(i)$ $\dots$
and $x\sin \theta -(1+\cos \theta )y=-a\sin \theta \dots ....(ii)$
Subtracting Eq. (ii) from Eq. (i), we get
$\begin{array}{ll}& (1-\cos \theta )y+(1+\cos \theta )y=2a\sin \theta \\ \Rightarrow & y[1-\cos \theta +1+\cos \theta ]=2a\sin \theta \\ \Rightarrow & y=a\sin \theta \end{array}$
Putting the value of $y$ in Eq. (i), we get $x\sin \theta +(1-\cos \theta )a\sin \theta =a\sin \theta$
$\begin{array}{ll}\Rightarrow & \sin \theta [x+(1-\cos \theta )a]=a\sin \theta \\ \Rightarrow & x+a-a\cos \theta =a\\ \Rightarrow & x-a\cos \theta =0\Rightarrow x=a\cos \theta \\ \text{ Now, }& x^2+y^2=(a\cos \theta {)}^2+(a\sin \theta {)}^2\\ \Rightarrow & x^2+y^2=a^2{\cos }^2\theta +a^2{\sin }^2\theta \\ \Rightarrow x^2+y^2& =a^2\left({\cos }^2\theta +{\sin }^2\theta \right)\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\end{array}$
$\Rightarrow x^2+y^2=a^2$, whose represent a circle.