If a point $(x, y)=( an heta+\sin heta, an heta-\sin heta)$, then the locus of $(x, y)$ is

Question

If a point $(x, y)=(	an 	heta+\sin 	heta, 	an 	heta-\sin 	heta)$, then the locus of $(x, y)$ is

Aicharya · Accepted Answer

Given that, $ \begin{array}{cc} & 	an 	heta+\sin 	heta=x \ & 	an 	heta-\sin 	heta=y \ & 	an 	heta=\frac{x+y}{2}, \sin 	heta=\frac{x-y}{2} \ 	herefore \quad & \left(\frac{x+y}{2}ight)\left(\frac{x-y}{2}ight)=	an 	heta \sin 	heta \ \Rightarrow \quad & x^2-y^2=4\left(\frac{\sin ^2 	heta}{\cos 	heta}ight)=4\left(\frac{1-\cos ^2 	heta}{\cos 	heta}ight) \ \Rightarrow \quad & x^2-y^2=4(\sec 	heta-\cos 	heta) \ \Rightarrow \quad & \left(x^2-y^2ight)^2=16\left(\sec 	heta^2+\cos ^2 	heta-2ight) \ \Rightarrow \quad & \left(x^2-y^2ight)^2=16\left(1+	an ^2 	heta+1-\sin ^2 	heta-2ight) \ \Rightarrow \quad & \left(x^2-y^2ight)^2=16\left(	an { }^2 	heta-\sin ^2 	hetaight) \ & \quad\left(x^2-y^2ight)^2=16 x y \end{array} $

If a point $(x, y)=(\tan \theta+\sin \theta, \tan \theta-\sin \theta)$, then the locus of $(x, y)$ is

Solution