Let $x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$ and

$y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$.

Then $\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$ at $t=\frac{\pi}{4}$ is equal to :

<p>$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t}$</p> <p>$\therefore$ $${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$$</p> <p>$\therefore$ $${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)$$</p> <p>and $${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$$</p> <p>$$\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)$$</p> <p>$\therefore$ $${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$$</p>