If x = 2 sqrt{2} sqrt{cos 2 theta} and y = 2 | vedclass

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(B) Given $x = 2 \sqrt{2} \sqrt{\cos 2 	heta}$ and $y = 2 \sqrt{2} \sqrt{\sin 2 	heta}$.Squaring both equations, we get $x^2 = 8 \cos 2 	heta$ and

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$-1$

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(B) Given $x = 2 \sqrt{2} \sqrt{\cos 2 	heta}$ and $y = 2 \sqrt{2} \sqrt{\sin 2 	heta}$.Squaring both equations, we get $x^2 = 8 \cos 2 	heta$ and $y^2 = 8 \sin 2 	heta$.Adding the squares: $x^2 + y^2 = 8(\cos 2 	heta + \sin 2 	heta)$.Alternatively, differentiate with respect to $	heta$:$\frac{dx}{d 	heta} = 2 \sqrt{2} \cdot \frac{1}{2 \sqrt{\cos 2 	heta}} \cdot (-2 \sin 2 	heta) = -\frac{2 \sqrt{2} \sin 2 	heta}{\sqrt{\cos 2 	heta}}$.$\frac{dy}{d 	heta} = 2 \sqrt{2} \cdot \frac{1}{2 \sqrt{\sin 2 	heta}} \cdot (2 \cos 2 	heta) = \frac{2 \sqrt{2} \cos 2 	heta}{\sqrt{\sin 2 	heta}}$.Then $\frac{dy}{dx} = \frac{dy/d 	heta}{dx/d 	heta} = \frac{2 \sqrt{2} \cos 2 	heta / \sqrt{\sin 2 	heta}}{-2 \sqrt{2} \sin 2 	heta / \sqrt{\cos 2 	heta}} = -\frac{\cos 2 	heta \sqrt{\cos 2 	heta}}{\sin 2 	heta \sqrt{\sin 2 	heta}} = -\cot 2 	heta \sqrt{\cot 2 	heta} = -(\cot 2 	heta)^{3/2}$.At $	heta = 22 \frac{1}{2}^{\circ} = \frac{45^{\circ}}{2}$, $2 	heta = 45^{\circ}$.So, $\frac{dy}{dx} = -(\cot 45^{\circ})^{3/2} = -(1)^{3/2} = -1$.

If $x = 2 \sqrt{2} \sqrt{\cos 2 \theta}$ and $y = 2 \sqrt{2} \sqrt{\sin 2 \theta}$, $0 < \theta < \frac{\pi}{4}$, then the value of $\frac{dy}{dx}$ at $\theta = 22 \frac{1}{2}^{\circ}$ is

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