Question

If $x=a{\cos }^3\theta ,y=b{\sin }^3\theta$ then:

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

Answer 1

The correct option is C

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

Explanation for the correct option.

$\begin{array}{rcl}x& =& a{\cos }^3\theta \\ & \Rightarrow & \frac{x}{a}={\cos }^3\theta \end{array}$

$$\begin{gathered} y=b{\sin }^3\theta \\ \Rightarrow \frac{y}{b}={\sin }^3\theta \end{gathered}$$

$$\begin{gathered} {\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3} \\ ={\left({\cos }^3\theta \right)}^{2/3}+{\left({\sin }^3\theta \right)}^{2/3} \\ ={\cos }^2\theta +{\sin }^2\theta \\ =1 \end{gathered}$$

Hence, option C is correct.