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 :

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Answer: D. \frac{-2}{3} x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} 	herefore {{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }} 	herefore {{dy} \over {dx}} = 	an 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} ight) 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}} \left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} ight) 	herefore {{1 + {{\left( {{{dy} \over {dx}}} ight)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}

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 :

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