Mathematics - Application of Derivative Question with Solution | TestHub

Let $\left(x_1,y_1\right)$ be one of the points of contact. Given curve is $y=\cos x$ $$\begin{array}{l}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}\\ {\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}|}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}=-\sin {\mathrm{x}}_1\end{array}$$ Now the equation of the tangent at $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is $$\begin{array}{rl}& y-y_1{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\left(x-x_1\right)\\ \Rightarrow & y-y_1=-\sin x_1\left(0-x_1\right)\end{array}$$ 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 $y=\cos x$ $$\therefore y_1=\cos x_1$$ From Eqs. (i), (ii), we get $$\begin{array}{rl}& {\sin }^2{x}_1+{\cos }^2{x}_1=\frac{y_1^2}{x_1^2}+y_1^2=1\\ \Rightarrow & x_1^2=y_1^2+y_1^2{x}_1^2\end{array}$$ Hence, the locus of $\left(x_1,y_1\right)$ is $$x^2=y^2+y^2{x}^2\Rightarrow x^2{y}^2=x^2-y^2$$