Question

Let C be a curve defined parametrically as $x=a{\cos }^3\theta ,y=a{\sin }^3\theta ,0\le \theta \le \frac{\mathrm{\pi }}{2}$. Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a). [CBSE 2014]

Answer 1

$$\begin{gathered} \mathrm{As},x=a{\cos }^3\theta \\ \Rightarrow \frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta \\ \mathrm{And},y=a{\sin }^3\theta \\ \Rightarrow \frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta \\ \mathrm{So},\frac{dy}{dx}=\frac{\left({\displaystyle \frac{dy}{d\theta }}\right)}{\left({\displaystyle \frac{dx}{d\theta }}\right)}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta \\ \mathrm{For}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{be}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{chord}\mathrm{joining}\mathrm{the}\mathrm{points}\left(a,0\right)\mathrm{and}\left(0,a\right), \\ \frac{dy}{dx}=\frac{0-a}{a-0} \\ \Rightarrow -\tan \theta =-1 \\ \Rightarrow \tan \theta =1 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{4} \\ \mathrm{Now}, \\ x=a{\cos }^3\frac{\mathrm{\pi }}{4}=a{\left(\frac{1}{\sqrt{2}}\right)}^3=\frac{a}{2\sqrt{2}}\mathrm{and} \\ y=a{\sin }^3\frac{\mathrm{\pi }}{4}=a{\left(\frac{1}{\sqrt{2}}\right)}^3=\frac{a}{2\sqrt{2}} \\ \mathrm{So},\mathrm{the}\mathrm{point}\mathrm{P}\mathrm{on}\mathrm{the}\mathrm{curve}C\mathrm{is}\left(\frac{\mathrm{a}}{2\sqrt{2}},\frac{\mathrm{a}}{2\sqrt{2}}\right). \end{gathered}$$