The method of eliminating `'theta'` from two given equations involving trigonometrical functions of `'theta'`. By using given equations involving `'theta'` and trigonometrical identities, we shall obtain an equation not involving `'theta'`.
On the basis of above information answer the following questions.
If `x sin^(3) theta+ y cos^(3)theta= sin theta cos theta` and `x sin theta- y cos theta = 0 `then (x, y) lie one

Options

A. a circle
B. a parabola
C. an ellipse
D. a hyperbola

Answer

To eliminate \( \theta \) from the given equations and find the relationship between \( x \) and \( y \), we can follow these steps:

### Step 1: Write down the given equations
The two equations provided are:
1. \( x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \) (Equation 1)
2. \( x \sin \theta - y \cos \theta = 0 \) (Equation 2)

### Step 2: Solve Equation 2 for one variable
From Equation 2, we can express \( x \) in terms of \( y \):
\[ x \sin \theta = y \cos \theta \implies \frac{x}{y} = \frac{\cos \theta}{\sin \theta} \implies \tan \theta = \frac{y}{x} \]
Thus, we have:
\[ \cot \theta = \frac{x}{y} \]

### Step 3: Substitute in Equation 1
Now, we will substitute \( \sin \theta \) and \( \cos \theta \) in terms of \( x \) and \( y \) into Equation 1.
We can divide both sides of Equation 1 by \( \sin \theta \cos \theta \):
\[ \frac{x \sin^3 \theta}{\sin \theta \cos \theta} + \frac{y \cos^3 \theta}{\sin \theta \cos \theta} = 1 \]
This simplifies to:
\[ x \frac{\sin^2 \theta}{\cos \theta} + y \frac{\cos^2 \theta}{\sin \theta} = 1 \]

### Step 4: Express \( \sin \theta \) and \( \cos \theta \)
Using the identities:
\[ \tan \theta = \frac{y}{x} \quad \text{and} \quad \cot \theta = \frac{x}{y} \]
We can express:
\[ \sin \theta = \frac{y}{\sqrt{x^2 + y^2}}, \quad \cos \theta = \frac{x}{\sqrt{x^2 + y^2}} \]

### Step 5: Substitute back into the equation
Substituting these into the equation gives:
\[ x \frac{y^2}{x^2} + y \frac{x^2}{y^2} = 1 \]
This simplifies to:
\[ \frac{xy^2}{x^2} + \frac{yx^2}{y^2} = 1 \]

### Step 6: Combine and simplify
Multiplying through by \( x^2y^2 \) to eliminate the denominators:
\[ y^2y^2 + x^2x^2 = x^2y^2 \]
This leads to:
\[ y^4 + x^4 = x^2y^2 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 = 1 \]

### Conclusion
The final equation \( x^2 + y^2 = 1 \) represents a circle with radius 1 centered at the origin.