Let x t = 2 2 cos t sin 2 t and y t = 2 2 sin t sin 2 t , t ∈ 0 , π 2 . Then 1 + d y d x 2 d 2 y…

Given

$x=2\sqrt{2}\cos t\sqrt{\sin 2t}$

Now differentiating w.r.t $t$ both side we get,

$\frac{dx}{dt}=\frac{2\sqrt{2}\cos 3\mathrm{t}}{\sqrt{\sin 2t}} ......\left(1\right)$

Also given $y\left(t\right)=2\sqrt{2}\sin t\sqrt{\sin 2t}$

Again differentiating w.r.t $t$ both side we get,

$\frac{dy}{dt}=\frac{2\sqrt{2}\sin 3t}{\sqrt{\sin 2t}} ......\left(2\right)$

Now dividing equation $\left(2\right)$ from $\left(1\right)$ we get,

$\frac{dy}{dx}=\tan 3t$

Now finding the value of $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$ we get,

$\frac{dy}{dx}=-1$

Now finding $\frac{d^{2}y}{d{x}^2}$ we get,

$\frac{d^{2}y}{d{x}^2}=\frac{3}{2\sqrt{2}}\frac{{\sec }^{2}3t·\sqrt{\sin 2t}}{\cos 3t}$

Now finding value of $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{\pi }{4}$ we get,

$\frac{d^{2}y}{d{x}^2}=-3$

Now putting the value of $\frac{dy}{dx} \& \frac{d^{2}y}{d{x}^2}$ in$\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}$ we get,

$\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{1+1}{-3}=-\frac{2}{3}$