Question

how to eliminate theta from x=acos 3 theta and y=bsin 3 theta

Answer 1

Answer:

hope this could help you

Answer 2

Answer:

$\implies \big(\frac{x}{a}\big)^{\frac{2}{3}}+\big(\frac{y}{b}\big)^{\frac{2}{3}} = 1$

Step-by-step explanation:

$Given \: x = acos^{3}\theta$

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

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

$\implies \left(\frac{x}{a}\right)^{\frac{1}{3}} = cos\theta\:---(1)$

$Given \: y = bsin^{3}\theta$

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

$\implies \left(\frac{y}{b}\right)^{\frac{1}{3}}=(sin^{3}\theta)^{\frac{1}{3}}$

$\implies \left(\frac{y}{b}\right)^{\frac{1}{3}} = sin\theta\:---(2)$

By Trigonometric identity:

$\boxed {\pink { cos^{2}\theta + sin^{2}\theta = 1}}$

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

$\implies \big(\frac{x}{a}\big)^{\frac{2}{3}}+\big(\frac{y}{b}\big)^{\frac{2}{3}} = 1$

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