#### Question

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

#### Solution

\[\text { As,} x = \text { a }\cos^3 \theta\]

\[ \Rightarrow \frac{dx}{d\theta} = - 3a \cos^2 \theta\sin\theta\]

\[\text { And }, y = a \sin^3 \theta\]

\[ \Rightarrow \frac{dy}{d\theta} = 3a \sin^2 \theta\cos\theta\]

\[So, \frac{dy}{dx} = \frac{\left( \frac{dy}{d\theta} \right)}{\left( \frac{dx}{d\theta} \right)} = \frac{3a \sin^2 \theta\cos\theta}{- 3a \cos^2 \theta\sin\theta} = - \tan\theta\]

\[\text { For the tangent to be parallel to the chord joining the points } \left( a, 0 \right) \text { and } \left( 0, a \right), \]

\[\frac{dy}{dx} = \frac{0 - a}{a - 0}\]

\[ \Rightarrow - \tan\theta = - 1\]

\[ \Rightarrow \tan\theta = 1\]

\[ \Rightarrow \theta = \frac{\pi}{4}\]

\[\text { Now }, \]

\[x = a \cos^3 \frac{\pi}{4} = a \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{a}{2\sqrt{2}} \text { and }\]

\[y = a \sin^3 \frac{\pi}{4} = a \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{a}{2\sqrt{2}}\]

\[\text { So, the point P on the curve C is } \left( \frac{a}{2\sqrt{2}}, \frac{a}{2\sqrt{2}} \right) .\]