Question

Prove that sum of intercepts of the tangent at any point to the curve represented by $x=3{\cos }^4\theta$ and $y=3{\sin }^4\theta$ on the co-ordinate axis is constant.

Answer 1

We have,

$x=3{\cos }^4\theta$

$y=3{\sin }^4\theta$

Now,

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

So,

$\frac{dy}{d\theta }=12{\sin }^3\theta \times \cos \theta$

$\frac{dx}{d\theta }=12{\cos }^3\theta \times (-\sin \theta )$

Now,

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{12{\sin }^3\theta \times \cos \theta }{12{\cos }^3\theta \times (-\sin \theta )}$

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

$\frac{dy}{dx}=-{\tan }^2\theta$

Equation of tangents

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

$\Rightarrow y-3{\sin }^4\theta =-\frac{{\sin }^2\theta }{{\cos }^2\theta }(x-3{\cos }^4\theta )$

$\Rightarrow y{\cos }^2\theta -3{\sin }^4\theta {\cos }^2\theta =-x{\sin }^2\theta +3{\cos }^4\theta {\sin }^2\theta$

$\Rightarrow x{\sin }^2\theta +y{\cos }^2\theta =3{\cos }^4\theta {\sin }^2\theta +3{\sin }^4\theta {\cos }^2\theta$

$\Rightarrow x{\sin }^2\theta +y{\cos }^2\theta =3{\cos }^2\theta {\sin }^2\theta ({\cos }^2\theta +{\sin }^2\theta )$

$\Rightarrow x{\sin }^2\theta +y{\cos }^2\theta =3{\cos }^2\theta {\sin }^2\theta \times 1$

$\Rightarrow x{\sin }^2\theta +y{\cos }^2\theta =3{\cos }^2\theta {\sin }^2\theta$

$\Rightarrow \frac{x{\sin }^2\theta +y{\cos }^2\theta }{3{\cos }^2\theta {\sin }^2\theta }=1$

$\Rightarrow \frac{x{\sin }^2\theta }{3{\cos }^2\theta {\sin }^2\theta }+\frac{y{\cos }^2\theta }{3{\cos }^2\theta {\sin }^2\theta }=1$

$\Rightarrow \frac{x}{3{\cos }^2\theta }+\frac{y}{3{\sin }^2\theta }=1$

Hence, proved.