Tangents are drawn to the curve y=sinx from the origin. The points of contact lie on

Let $\mathrm{P}\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)$ Lies on the curve $\mathrm{y}=\sin \mathrm{x}$

                    ${\mathrm{y}}_1=\sin {\mathrm{x}}_1 -\left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\cos \mathrm{x} \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)}=\cos {\mathrm{x}}_1$

eq of Tangent $\mathrm{y}-{\mathrm{y}}_1=\left(\cos \left({\mathrm{x}}_1\right)\right) \left(\mathrm{x}-{\mathrm{x}}_1\right)$

its passes through (0, 0)

$\begin{array}{rcl} & \Rightarrow & \left(-{\mathrm{y}}_1\right)=\left(\cos {\mathrm{x}}_1\right) \left(0-{\mathrm{x}}_1\right)\\ & \Rightarrow & \cos {\mathrm{x}}_1=\frac{{\mathrm{y}}_1}{{\mathrm{x}}_1} -\left(2\right)\end{array}$

from eq (1) & eq (2)

${{\mathrm{y}}_1}^2+\frac{{{\mathrm{y}}_1}^2}{{{\mathrm{x}}_1}^2} \Rightarrow$ Locus of $\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)$ is ${\mathrm{y}}^2+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}=1$

                     $\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2+{\mathrm{y}}^2={\mathrm{x}}^2$