UPSEE 2017 Solved Paper

Given, $y=x^{2}$ ......(i) $x=\frac{5}{3}\cos t$, $y=\frac{5}{4}\sin t$ .......(ii) Which is parametric equation, we change this equation is cartesian equation as follows: $\cos t=\frac{3}{5}x, \sin t=\frac{4}{5}y$ On squaring and adding both i.e. $\cos t$ and sin t, we get $\frac{9}{25}x^{2}+\frac{16}{25}y^{2}=\cos^{2}t+\sin^{2}t$ $\Rightarrow 9x^{2}+16y^{2}=25$ ........(iii) $[\because \cos^{2}\theta+\sin^{2}\theta=1]$ -The intersection points at Eq. (i) and (iii) are $(1,1)$ and $(-1,1)$. Now, slope of tangent of Eq. (i) at point $(1,1)$ is $m_{1}=\frac{dy}{dx}=2x$ $\therefore$ $m_{1}=\left.\frac{dy}{dx}\right|_{(1,1)}=2$ And slope of tangent of Eq. (iii), at point $(1,1)$ is $m_{2}=\frac{dy}{dx}=-\frac{9}{16}$ $\therefore$ Angle at point of intersection of Eqs. (i) and (iii), we get $\theta_{1}=\tan^{-1}\left|\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|$ $\theta_{1}=\tan^{-1}\left|\frac{2+\frac{9}{16}}{1-\frac{2\times9}{16}}\right|$ $=\tan^{-1}\frac{41}{2}$ Similarly, slope of tangent of Eq. (i) at point $(-1,1)$ $m_{1}=\left.\frac{dy}{dx}\right|_{(-1,1)}=-2$ And slope of tangent of Eq. (iii) at point $(-1,1)$ $m_{2}=\frac{dy}{dx}=-\frac{9}{16}$ $\therefore$ Angle at point of intersection of Eqs. (i) and (iii), we get $\theta_{2}=\tan^{-1}\left|\frac{\frac{-2-9}{16}}{1-\frac{18}{16}}\right|$ $=\tan^{-1}\frac{41}{2}$