Question

Find the equation of the tangent and normal to the curve $x={\sin }^3\theta$ and $y=aco{s}^3\theta$ at $\theta =\frac{\pi }{4}$.

Answer 1

$x={\sin }^3\theta ,y=a{\cos }^3\theta$

$x={\sin }^3\theta \to \frac{dx}{d\theta }=3{\sin }^2\theta \cos \theta$

$y=a{\cos }^3\theta \to \frac{dy}{d\theta }=-3a{\cos }^2\theta \sin \theta$

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-3a{\cos }^2\theta \sin \theta }{3{\sin }^2\theta \cos \theta }=-a\frac{\cos \theta }{\sin \theta }$

Equation of tangent,

$y-y_1=\frac{dy}{dx}(x-x_1)$

$y-a{\cos }^3\theta =(-a\frac{\cos \theta }{\sin \theta })(x-{\sin }^3\theta )$

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

$y-\left(\frac{a}{2\sqrt{2}}\right)=(-a)(x-\frac{1}{2\sqrt{2}})$

$y-\left(\frac{a}{2\sqrt{2}}\right)=-ax+\left(\frac{a}{2\sqrt{2}}\right)$

$ax+y=\left(\frac{a}{\sqrt{2}}\right)$

$\sqrt{2}ax+\sqrt{2}y=a$

Equation of normal,

$y-a{\cos }^3\theta =(-\frac{\sin \theta }{a\cos \theta })(x-{\sin }^3\theta )$

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

$y-\left(\frac{a}{2\sqrt{2}}\right)=-\frac{1}{a}(x-\frac{1}{2\sqrt{2}})$

$y-\left(\frac{a}{2\sqrt{2}}\right)=-\frac{x}{a}+\frac{1}{a2\sqrt{2}}$

$y+\frac{x}{a}=\frac{1}{a2\sqrt{2}}+\frac{a}{2\sqrt{2}}$

$y+\frac{x}{a}=\frac{a^2+1}{a2\sqrt{2}}$

$2\sqrt{2}x+a2\sqrt{2}y=a^2+1$