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

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

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

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

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

x=a \cos ^3 \theta, y=b \sin ^3 \theta

, then

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is derived by expressing

\cos \theta

and

\sin \theta

in terms of x, y, a, and b, and then using the identity

\sin^2 \theta + \cos^2 \theta = 1