1. Given parametric equations

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

We need to find

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

at $t=\frac{\pi}{4}$.

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2. Simplify the parametric form

Since

$\sin 2t = 2\sin t\cos t,$

we can first observe a useful relation:

$x^2+y^2=(2\sqrt2)^2(\cos^2 t+\sin^2 t)(\sin 2t)=8\sin 2t.$

Also,

$\frac{y}{x}=\tan t.$

But an even better simplification comes from squaring:

$x=2\sqrt2\cos t\sqrt{\sin 2t}$

so

$x^2=8\cos^2 t\sin 2t=8\cos^2 t(2\sin t\cos t)=16\sin t\cos^3 t.$

Similarly,

$y^2=16\sin^3 t\cos t.$

Hence,

$\sqrt{x^2y^2}=xy=16\sin^2 t\cos^2 t=4\sin^2 2t.$

But the cleanest route is to rewrite using

$r=2\sqrt2\sqrt{\sin 2t}, \quad x=r\cos t, \quad y=r\sin t.$

At $t=\frac\pi4$, we can compute derivatives directly.

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3. Differentiate $x(t)$ and $y(t)$

Let

$s=\sqrt{\sin 2t}=(\sin 2t)^{1/2}.$

Then

$x=2\sqrt2\cos t\,s, \qquad y=2\sqrt2\sin t\,s.$

Now,

$\frac{ds}{dt}=\frac{1}{2}(\sin 2t)^{-1/2}(2\cos 2t)=\frac{\cos 2t}{\sqrt{\sin 2t}}=\frac{\cos 2t}{s}.$

So,

$\frac{dx}{dt}=2\sqrt2\left(-\sin t\,s+\cos t\,\frac{\cos 2t}{s}\right),$

$\frac{dy}{dt}=2\sqrt2\left(\cos t\,s+\sin t\,\frac{\cos 2t}{s}\right).$

Therefore,

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$

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4. Evaluate at $t=\frac\pi4$

At $t=\frac\pi4$,

$\sin t=\cos t=\frac{1}{\sqrt2}, \qquad \sin 2t=1, \qquad \cos 2t=0, \qquad s=1.$

Thus,

$\frac{dx}{dt}=2\sqrt2\left(-\frac{1}{\sqrt2}\right)=-2,$

$\frac{dy}{dt}=2\sqrt2\left(\frac{1}{\sqrt2}\right)=2.$

Hence,

$\frac{dy}{dx}=\frac{2}{-2}=-1.$

So,

$1+\left(\frac{dy}{dx}\right)^2=1+1=2.$

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5. Find $\dfrac{d^2y}{dx^2}$

For parametric curves,

$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}.$

So we need $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ at $t=\frac\pi4$.

Let

$u=\frac{dy}{dt}, \qquad v=\frac{dx}{dt}, \qquad \frac{dy}{dx}=\frac{u}{v}.$

Then

$\frac{d}{dt}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}.$

So we need $u'$ and $v'$.

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6. Compute second derivatives with respect to $t$

We already have

$u=2\sqrt2\left(\cos t\,s+\sin t\,\frac{\cos 2t}{s}\right),$

$v=2\sqrt2\left(-\sin t\,s+\cos t\,\frac{\cos 2t}{s}\right).$

Instead of differentiating these fully, we use a simpler identity.

Let

$x=2\sqrt2\cos t\,s, \qquad y=2\sqrt2\sin t\,s.$

Then at $t=\frac\pi4$, since $\cos 2t=0$, the derivatives simplify a lot. We compute $x''(t)$ and $y''(t)$ there.

First,

$x'=2\sqrt2\left(-\sin t\,s+\cos t\,s'\right),$

$y'=2\sqrt2\left(\cos t\,s+\sin t\,s'\right),$

where

$s'=\frac{\cos 2t}{s}.$

Differentiate again:

$x''=2\sqrt2\left(-\cos t\,s-2\sin t\,s'+\cos t\,s''\right),$

$y''=2\sqrt2\left(-\sin t\,s+2\cos t\,s'+\sin t\,s''\right).$

Now compute $s''$ at $t=\frac\pi4$.

Since

$s' = \cos 2t\,(\sin 2t)^{-1/2},$

$s'' = -2\sin 2t\,(\sin 2t)^{-1/2}+\cos 2t\left(-\frac12\right)(\sin 2t)^{-3/2}(2\cos 2t).$

At $t=\frac\pi4$,

$\sin 2t=1, \quad \cos 2t=0,$

so

$s''=-2.$

Therefore,

=2\sqrt2\left(-\frac3{\sqrt2}\right)=-6,$$ $$y''=2\sqrt2\left(-\frac1{\sqrt2}(1)+0+\frac1{\sqrt2}(-2)\right) =2\sqrt2\left(-\frac3{\sqrt2}\right)=-6.$$ So at $t=\frac\pi4$, $$u=y'=2, \quad v=x'=-2, \quad u'=y''=-6, \quad v'=x''=-6.$$ Thus, $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{u'v-uv'}{v^2} =\frac{(-6)(-2)-2(-6)}{(-2)^2} =\frac{12+12}{4}=6.$$ Hence, $$\frac{d^2y}{dx^2}=\frac{6}{dx/dt}=\frac{6}{-2}=-3.$$ --- 7. **Compute the required quantity** $$\frac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}} =\frac{2}{-3}=-\frac23.$$ --- 8. **Match with options** The correct option is $$\boxed{-\frac23}$$ which is **Option D**. --- 9. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So they agree.