TS EAMCET 5 May 2018 Shift 1 Solved Paper

Assume the point of intersection be $(x_{0}, y_{0})$ $x^{2}-y^{2}=4$ Differentiate the above equation with respect to $x$. $2 x-2 y \frac{d y}{d x}=0$ $\frac{d y}{d x}=\frac{x}{y}$ At point $(x_{0}, y_{0})$ $m_{1}=\frac{x_{0}}{y_{0}}$ and curve $x^{2}+y^{2}=4\sqrt{2}$ Differentiate with respect to $x$ $2 x+2 y \frac{d y}{d x}=0$ $\frac{d y}{d x}=\frac{x}{y}$ At point $(x_{0}, y_{0})$ $m_{2}=-\frac{x_{0}}{y_{0}}$ The angle between the tangents is calculated as, $\theta = \tan^{-1} \left| \frac{ \frac{x_0}{y_0} + \frac{x_0}{y_0} }{ 1+ \frac{x_0}{y_0} \left( -\frac{x_0}{y_0} \right) } \right|$ $=\tan^{-1} \left( \frac{x_0 y_0}{2} \right)$ since, the point $(x_{0}, y_{0})$ lies on curves then, $x_{0}^{2}-y_{0}^{2}=4$ and $x_{0}^{2}+y_{0}^{2}=4\sqrt{2}$ Then, $x_{0} y_{0}=2$ …… (2) Substitute the value in equation (1), $\theta = \tan^{-1} \left( \frac{2}{2} \right)$ $=\frac{\pi}{4}$