TS EAMCET 2017 Code A Solved Paper

Given equation of the curves are, $x^{2}=8 y \ldots (1)$ $x y=8 \ldots (2)$ On solving equation (1) and $(2),$ we get $x=4$ $y=2$ Therefore, the point of intersection of given curves is (4,2). Slope of tangent to the curve (1) at point (4,2) is calculated as $x^{2}=8 y$ $2x=8 \;\; \frac{dy}{dx} \Rightarrow \;\; \frac{dy}{dy}=\;\; \frac{x}{4}$ $\Rightarrow m_{1}=1$ Slope of tangent to the curve ( 2 ) at point (4,2) is calculated $a s$ $xy=8$ $x \;\; \frac{dy}{dx} + y = 0 \Rightarrow \;\; \frac{dy}{dx} = -\;\; \frac{y}{x}$ $\Rightarrow m_{2}=-\;\; \frac{1}{2}$ The angle $\theta$ between the curves is calculated as, $\tan \theta = \left( \;\; \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} \right)$ $= \left( \;\; \frac{ \frac{-1}{2} - 1 }{ 1 - \frac{1}{2} } \right)$ $\tan \theta = -3$ $\theta = \tan^{-1}(-3)$