Question

the coordinates of a particle moving in a plane are given by x=4cos6t and y=6sin6t.find the equation of the path of the particle​

Answer 1

Answer:

$\bf{(\frac{x}{4})^{\frac{1}{3}}+(\frac{y}{6})^{\frac{1}{3}}=1}$

Step-by-step explanation:

$\text{Given:}$

$x=4\:cos^6t\:and\:y=6\:sin^6t$

$\implies\:\frac{x}{4}=cos^6t\:and\:\frac{y}{6}=sin^6t$

$\implies\:(\frac{x}{4})^{\frac{1}{3}}=(cos^6t)^{\frac{1}{3}}\:and\:(\frac{y}{6})^{\frac{1}{3}}=(sin^6t)^{\frac{1}{3}}$

$\implies\:(\frac{x}{4})^{\frac{1}{3}}=cos^2t\:and\:(\frac{y}{6})^{\frac{1}{3}}=sin^2t$

Adding these equations

$(\frac{x}{4})^{\frac{1}{3}}+(\frac{y}{6})^{\frac{1}{3}}=cos^2t+sin^2t$

$\implies\:\bf{(\frac{x}{4})^{\frac{1}{3}}+(\frac{y}{6})^{\frac{1}{3}}=1}$