Tangents are drawn from the origin to the curve y= x. Their points...

Let $\left(x_{1}, y_{1}\right)$ be one of the points of contact. Given curve is $y=\cos x$ $$\begin{array}{l} \Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}\\ \left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}=-\sin \mathrm{x}_{1} \end{array}$$Now the equation of the tangent at \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$is$$\begin{aligned}& y-y_{1}\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}\left(x-x_{1}\right) \\ \Rightarrow & y-y_{1}=-\sin x_{1}\left(0-x_{1}\right) \end{aligned}$$Since, it is given that equation of tangent passes through origin.$$\begin{array}{lr} \therefore& 0-y_{1}=-\sin x_{1}\left(0-x_{1}\right) \\ \Rightarrow & y_{1}=-x_{1} \sin x_{1} \end{array}$$Also, point$\left(x_{1}, y_{1}\right)$lies on$y=\cos x$\therefore \quad y_{1}=\cos x_{1} From Eqs. (i), (ii), we get$$\begin{aligned}& \sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1 \\ \Rightarrow & x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2} \end{aligned}$$Hence, the locus of$\left(x_{1}, y_{1}\right)$ is x^{2}=y^{2}+y^{2} x^{2} \Rightarrow x^{2} y^{2}=x^{2}-y^{2}