Question

The equation of the normal to the curve x = a cos³ θ, y = a sin³ θ at the point θ = π/4 is

(a) x = 0
(b) y = 0
(c) c = y
(d) x + y = a

Answer 1

(c) x = y

$$\begin{gathered} \mathrm{Here}, \\ x=a{\cos }^3\theta \mathrm{and}y=a{\sin }^3\theta \\ \frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta \mathrm{and}\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta \\ \mathrm{Now}, \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{4}}\text{=-tan}\frac{\mathrm{\pi }}{4}\text{=-1} \\ \left(x_1,y_1\right)=\left(a{\cos }^3\frac{\mathrm{\pi }}{4},a{\sin }^3\frac{\mathrm{\pi }}{4}\right)=\left(\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}}\right) \\ \therefore \text{Equation of the normal} \\ =y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-\frac{a}{2\sqrt{2}}=1\left(x-\frac{a}{2\sqrt{2}}\right) \\ \Rightarrow x=y \end{gathered}$$