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>
$x \sin \theta+(1-\cos \theta) y=a \sin \theta .... (i)$ $\ldots$
and $x \sin \theta-(1+\cos \theta) y=-a \sin \theta \quad \ldots .... (ii)$
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 $x \sin \theta+(1-\cos \theta) a \sin \theta=a \sin \theta$
$\begin{array}{ll}\Rightarrow & \sin \theta[x+(1-\cos \theta) a]=a \sin \theta \\ \Rightarrow & x+a-a \cos \theta=a \\ \Rightarrow & x-a \cos \theta=0 \Rightarrow x=a \cos \theta \\ \text { Now, } & x^{2}+y^{2}=(a \cos \theta)^{2}+(a \sin \theta)^{2} \\ \Rightarrow & x^{2}+y^{2}=a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta \\ \Rightarrow x^{2}+y^{2} & =a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]\end{array}$
$\Rightarrow x^{2}+y^{2}=a^{2}$, whose represent a circle.