Question

Eliminate $\theta$ if, $x=a{\cos }^3\theta$, $y=b{\sin }^3\theta$.

Answer 1

consider given equation,

$x=a{\cos }^3\theta ......\left(1\right)andy=a{\sin }^3\theta ......\left(2\right)$

From equation (1) and (2) to,

$\frac{x}{a}={\cos }^3\theta and\frac{y}{a}={\sin }^3\theta$

${\left(\frac{x}{a}\right)}^{\frac{1}{3}}=\cos \theta and{\left(\frac{y}{a}\right)}^{\frac{1}{3}}=\sin \theta$

Then,

$\cos \theta ={\left(\frac{x}{a}\right)}^{\frac{1}{3}}......\left(3\right)and\sin \theta ={\left(\frac{y}{a}\right)}^{\frac{1}{3}}......\left(4\right)]$

Squaring both side equation (3) and (4), we get

${\cos }^2\theta ={\left(\frac{x}{a}\right)}^{\frac{2}{3}}......\left(5\right)and{\sin }^2\theta ={\left(\frac{y}{a}\right)}^{\frac{2}{3}}......\left(6\right)]$

Adding equation (5) and (6) to,

${\cos }^2\theta +{\sin }^2\theta ={\left(\frac{x}{a}\right)}^{\frac{2}{3}}+{\left(\frac{y}{a}\right)}^{\frac{2}{3}}]$

$[1={\left(\frac{x}{a}\right)}^{\frac{2}{3}}+{\left(\frac{y}{a}\right)}^{\frac{2}{3}}]$

$[{\left(\frac{x}{a}\right)}^{\frac{2}{3}}+{\left(\frac{y}{a}\right)}^{\frac{2}{3}}=1]$

The above equation is independent to $\theta$.