Find the slope of the normal to the curve x=a cos^3 θ , y= a sin^3 θ…

Slope of the tamgent of the curve is dy/dx

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

$sin^3\theta = \frac{3sin\theta - sin3\theta}{4}$

$ \frac{dy}{d\theta} = a(\frac{3cos\theta - 3cos3\theta}{4})$

$cos^3\theta = \frac{3cos\theta + cos3\theta}{4}$

$ \frac{dx}{d\theta} =a( -\frac{3sin\theta - 3sin3\theta}{4})$

$\frac{dy}{dx} = \frac{3cos\theta - 3cos3\theta}{3sin\theta - 3sin3\theta}$

At $\theta = \frac{\pi}{4} $

$\text{Slope of the normal is }-\frac{dx}{dy} = -\frac{3sin\theta - 3sin3\theta}{3cos\theta - 3cos3\theta}=1$