Question

The locus of the point $(a{\cos }^3\theta ,b{\sin }^3\theta )$ where $0\le \theta <2\pi$ is

• A. $(x^{2}y{)}^{2/3}+(x{y}^2{)}^{2/3}=1$
• B. $(x^2{y}^2{)}^{2/3}+(x{y}^2{)}^{2/3}=1$
• C. $(x/a{)}^{2/3}+(y/b{)}^{2/3}=1$
• D. $(x^2/a{)}^{2/3}+(y^2/b{)}^{2/3}=1$

Answer 1

The correct option is C $(x/a{)}^{2/3}+(y/b{)}^{2/3}=1$

Given point $(x,y)=(a{\cos }^3\theta ,b{\sin }^3\theta )$

$\Rightarrow x=a{\cos }^3\theta$

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

${\left(\frac{x}{a}\right)}^{2/3}={\cos }^2\theta$

$\Rightarrow y=b{\sin }^3\theta$

$\frac{y}{b}={\sin }^3\theta$

${\left(\frac{y}{b}\right)}^{2/3}={\sin }^2\theta$

$\therefore {\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3}=1.$

Hence, the answer is ${\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3}=1.$