If x = a cos⁴ θ, y = a sin⁴ θ. then dy/dx at θ = 3π/4 is

The correct option is -1.

We are given the parametric equations:
$$x=a{\cos }^4\theta$$
$$y=a{\sin }^4\theta$$

To find the derivative $\frac{dy}{dx}$, we use the chain rule for parametric differentiation:
$$\frac{dy}{dx}=\frac{\left({\displaystyle \frac{dy}{d\theta }}\right)}{\left({\displaystyle \frac{dx}{d\theta }}\right)}$$

First, we differentiate x with respect to θ:
$$\frac{dx}{d\theta }=\frac{d}{d\theta }\left(a{\cos }^4\theta \right)=a\cdot 4{\cos }^3\theta \cdot (-\sin \theta )=-4a{\cos }^3\theta \sin \theta$$

Next, we differentiate y with respect to θ:
$$\frac{dy}{d\theta }=\frac{d}{d\theta }\left(a{\sin }^4\theta \right)=a\cdot 4{\sin }^3\theta \cdot \left(\cos \theta \right)=4a{\sin }^3\theta \cos \theta$$

Now, substitute these derivatives into the formula for $\frac{dy}{dx}$:
$$\frac{dy}{dx}=\frac{4a{\sin }^3\theta \cos \theta }{-4a{\cos }^3\theta \sin \theta }$$

By cancelling out the common factors $4a\sin \theta \cos \theta$ from the numerator and the denominator, we simplify the expression to:
$$\frac{dy}{dx}=-\frac{{\sin }^2\theta }{{\cos }^2\theta }=-{\tan }^2\theta$$

We need to evaluate $\frac{dy}{dx}$ at $\theta =\frac{3\pi }{4}$:
$${\left(\frac{dy}{dx}\right)}_{\theta =\frac{3\pi }{4}}=-{\tan }^2\left(\frac{3\pi }{4}\right)$$

Since $\tan \left(\frac{3\pi }{4}\right)=\tan (\pi -\frac{\pi }{4})=-\tan \left(\frac{\pi }{4}\right)=-1$, we square this value:
$${\left(\frac{dy}{dx}\right)}_{\theta =\frac{3\pi }{4}}=-{(-1)}^2=-\left(1\right)=-1$$

Thus, the value of $\frac{dy}{dx}$ at the given point is -1.