What is the eliminant of `x=a cos^3 theta, y=b sin^3 theta` ?

Answer

To find the eliminant of the equations \( x = a \cos^3 \theta \) and \( y = b \sin^3 \theta \), we will eliminate the parameter \( \theta \) from these equations. Here’s a step-by-step solution:

### Step 1: Rewrite the equations
We start with the given equations:
\[ x = a \cos^3 \theta \]
\[ y = b \sin^3 \theta \]

### Step 2: Express \( \cos \theta \) and \( \sin \theta \)
From the first equation, we can express \( \cos \theta \):
\[ \cos \theta = \left(\frac{x}{a}\right)^{\frac{1}{3}} \]
From the second equation, we can express \( \sin \theta \):
\[ \sin \theta = \left(\frac{y}{b}\right)^{\frac{1}{3}} \]

### Step 3: Square both expressions
Next, we square both expressions to find \( \cos^2 \theta \) and \( \sin^2 \theta \):
\[ \cos^2 \theta = \left(\frac{x}{a}\right)^{\frac{2}{3}} \]
\[ \sin^2 \theta = \left(\frac{y}{b}\right)^{\frac{2}{3}} \]

### Step 4: Use the Pythagorean identity
We know from trigonometry that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting our expressions for \( \cos^2 \theta \) and \( \sin^2 \theta \) into this identity gives:
\[ \left(\frac{x}{a}\right)^{\frac{2}{3}} + \left(\frac{y}{b}\right)^{\frac{2}{3}} = 1 \]

### Step 5: Final form of the eliminant
Rearranging the equation, we get the eliminant:
\[ \frac{x^{\frac{2}{3}}}{a^{\frac{2}{3}}} + \frac{y^{\frac{2}{3}}}{b^{\frac{2}{3}}} = 1 \]

### Conclusion
Thus, the eliminant of the given equations is:
\[ \frac{x^{\frac{2}{3}}}{a^{\frac{2}{3}}} + \frac{y^{\frac{2}{3}}}{b^{\frac{2}{3}}} = 1 \]

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