If $x = a\sin 2t(1 + \cos 2t)$ and $y = b\cos 2t(1 - \cos 2t),$ then show that ${\left( {\frac{{dy}}{{dx}}} \right)_{t = \pi /4}} = \frac{b}{a}$

and $y = b\cos 2t(1 - \cos 2t)$
therefore,$\frac{{dx}}{{dt}} = a\left[ {\sin 2t \cdot \frac{d}{{dt}}(1 + \cos 2t) + (1 + \cos 2t) \cdot \frac{d}{{dt}}\sin 2t} \right]$

$= a\left[ {\sin 2t \cdot ( - \sin 2t) \cdot \frac{d}{{dt}}2t + (1 + \cos 2t) \cdot \cos 2t \cdot \frac{d}{{dt}}2t} \right]$
$= - 2a{\sin ^2}2t + 2a\cos 2t(1 + \cos 2t)$

$\Rightarrow$ $\frac{{dx}}{{dt}} = - 2a\left[ {{{\sin }^2}2t - \cos 2t(1 + \cos 2t)} \right]$ ……(i)

and $\frac{{dy}}{{dt}} = b\left[ {\cos 2t \cdot \frac{d}{{dt}}(1 - \cos 2t) + (1 - \cos 2t) \cdot \frac{d}{{dt}}\cos 2t} \right]$
$= b\left[ {\cos 2t \cdot (\sin 2t)\frac{d}{{dt}}2t + (1 - \cos 2t)( - \sin 2t) \cdot \frac{d}{{dt}}2t} \right]$

$= b[2\sin 2t \cdot \cos 2t + 2(1 - \cos 2t)( - \sin 2t)]$

$= 2b[\sin 2t \cdot \cos 2t - (1 - \cos 2t)\sin 2t]$ …….(ii)

therefore,$\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{ - 2b[ - \sin 2t \cdot \cos 2t + (1 - \cos 2t)\sin 2t]}}{{ - 2a\left[ {{{\sin }^2}2t - \cos 2t(1 + \cos 2t)} \right]}}$

$\Rightarrow$ ${\left( {\frac{{dy}}{{dx}}} \right)_{t = \pi /4}} = \frac{b}{a}\frac{{\left[ { - \sin \frac{\pi }{2}\cos \frac{\pi }{2} + \left( {1 - \cos \frac{\pi }{2}} \right)\sin \frac{\pi }{2}} \right]}}{{\left[ {{{\sin }^2}\frac{\pi }{2} - \cos \frac{\pi }{2}\left( {1 + \cos \frac{\pi }{2}} \right)} \right]}}$

$= \frac{b}{a} \cdot \frac{{(0 + 1)}}{{(1 - 0)}}$ and $\left. {\cos \frac{\pi }{2} = 0} \right]$
$= \frac{b}{a}$

Hence proved.