Let `cos^-1 (x/2)+cos^-1 (y/3)=theta` and denote by `f(x,y,theta)=0` the rational integeral expression in `x and y.` Then for `theta=pi/2` the locus represented by `f(x,y,pi/2)=0` is (A) an ellipse (B) parabola (C) hyperbola (D) pair of lines

Options

A. An ellipse
B. A parabola
C. A hyperbola
D. A pair of line

Answer

To solve the problem, we start with the equation given:

\[ \cos^{-1}\left(\frac{x}{2}\right) + \cos^{-1}\left(\frac{y}{3}\right) = \theta \]

We need to find the locus represented by \( f(x, y, \frac{\pi}{2}) = 0 \).

### Step 1: Set up the equations
Let:
\[ \cos^{-1}\left(\frac{x}{2}\right) = \alpha \quad \text{and} \quad \cos^{-1}\left(\frac{y}{3}\right) = \beta \]

Thus, we can rewrite the equation as:
\[ \alpha + \beta = \theta \]

### Step 2: Express \( x \) and \( y \) in terms of \( \alpha \) and \( \beta \)
From the definitions of \( \alpha \) and \( \beta \), we have:
\[ \frac{x}{2} = \cos(\alpha) \quad \Rightarrow \quad x = 2\cos(\alpha) \]
\[ \frac{y}{3} = \cos(\beta) \quad \Rightarrow \quad y = 3\cos(\beta) \]

### Step 3: Use the cosine addition formula
Using the cosine addition formula, we have:
\[ \cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \]
Substituting \( \theta \) into the equation, we get:
\[ \cos(\theta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \]

### Step 4: Substitute \( \cos(\alpha) \) and \( \cos(\beta) \)
Substituting the expressions for \( x \) and \( y \):
\[ \cos(\alpha) = \frac{x}{2}, \quad \cos(\beta) = \frac{y}{3} \]
Thus, we have:
\[ \cos(\theta) = \left(\frac{x}{2}\right)\left(\frac{y}{3}\right) - \sin(\alpha)\sin(\beta) \]

### Step 5: Find \( \sin(\alpha) \) and \( \sin(\beta) \)
Using the Pythagorean identity:
\[ \sin(\alpha) = \sqrt{1 - \cos^2(\alpha)} = \sqrt{1 - \left(\frac{x}{2}\right)^2} = \frac{\sqrt{4 - x^2}}{2} \]
\[ \sin(\beta) = \sqrt{1 - \cos^2(\beta)} = \sqrt{1 - \left(\frac{y}{3}\right)^2} = \frac{\sqrt{9 - y^2}}{3} \]

### Step 6: Substitute back into the equation
Substituting \( \sin(\alpha) \) and \( \sin(\beta) \) into the equation:
\[ \cos(\theta) = \frac{xy}{6} - \left(\frac{\sqrt{4 - x^2}}{2}\right)\left(\frac{\sqrt{9 - y^2}}{3}\right) \]

### Step 7: Set \( \theta = \frac{\pi}{2} \)
For \( \theta = \frac{\pi}{2} \), we know that \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ 0 = \frac{xy}{6} - \frac{\sqrt{4 - x^2}\sqrt{9 - y^2}}{6} \]

### Step 8: Rearranging the equation
This leads to:
\[ xy = \sqrt{(4 - x^2)(9 - y^2)} \]

### Step 9: Square both sides
Squaring both sides gives:
\[ x^2y^2 = (4 - x^2)(9 - y^2) \]
Expanding the right side:
\[ x^2y^2 = 36 - 9x^2 - 4y^2 + x^2y^2 \]
Cancelling \( x^2y^2 \) from both sides:
\[ 0 = 36 - 9x^2 - 4y^2 \]
Rearranging gives:
\[ 9x^2 + 4y^2 = 36 \]

### Step 10: Divide by 36
Dividing the entire equation by 36:
\[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \]

### Conclusion
This is the standard form of the equation of an ellipse.

Thus, the locus represented by \( f(x, y, \frac{\pi}{2}) = 0 \) is an ellipse.

### Final Answer
The correct option is (A) an ellipse.