Question

Prove that sum of intercepts of the tangent at and point to the curve represented by $x={cos}^4\theta \&y={sin}^4\theta$ on the coordinate axis is constant

Answer 1

We have,

$x={\cos }^4\theta$ ….. (1)

$y={\sin }^4\theta$ …… (2)

On differentiation (1) and (2) to, we get,

$\frac{dx}{d\theta }=-4{\cos }^3\theta \sin \theta$

$\frac{dy}{d\theta }=4{\sin }^3\theta \cos \theta$

Then,

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{4{\sin }^3\theta \cos \theta }{-4{\cos }^3\theta \sin \theta }$

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

Now equation of tangent is

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

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

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

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

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

$x{\sin }^2\theta +y{\cos }^2\theta ={\sin }^2\theta {\cos }^2\theta ({\cos }^2\theta +{\sin }^2\theta )\therefore {\cos }^2\theta +{\sin }^2\theta =1$

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

On divide both side ${\sin }^2\theta {\cos }^2\theta$

So,

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

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

$x{\sec }^2\theta +y{\csc }^2\theta =1$

It is equation of tangent

Hence, this is the answer.