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

Question

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

Aicharya · Accepted Answer

Given, equations of straight lines are

Subtracting Eq. (ii) from Eq. (i), we get $ \begin{array}{ll} & (1-\cos \theta) y+(1+\cos \theta) y=2 a \sin \theta \\ \Rightarrow & y[1-\cos \theta+1+\cos \theta]=2 a \sin \theta \\ \Rightarrow & y=a \sin \theta \end{array} $ Putting the value of $y$ in Eq. (i), we get $ \begin{array}{ll} & x \sin \theta+(1-\cos \theta) a \sin \theta=a \sin \theta \\ \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, } \quad & 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] \\ \Rightarrow & x^2+y^2=a^2, \text { whose represent a circle. } \end{array} $

For any value of $\theta$, if the straight lines $x \sin \theta+(1-\cos \theta) y=a \sin \theta$ and $x \sin…

Solution