If P & Q are the ends of a pair of conjugate diameters & C is the centre of the ellipse `4x^2+9y^2=36`. Then the area of `triangleCPQ` is:

Answer

To find the area of triangle CPQ where P and Q are the ends of a pair of conjugate diameters of the ellipse given by the equation \(4x^2 + 9y^2 = 36\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse
The given equation of the ellipse is:
\[ 4x^2 + 9y^2 = 36 \]
We can divide the entire equation by 36 to get it into standard form:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This shows that the semi-major axis \(a = 3\) and the semi-minor axis \(b = 2\).

### Step 2: Identify the center of the ellipse
The center \(C\) of the ellipse is at the origin:
\[ C(0, 0) \]

### Step 3: Determine the endpoints of the conjugate diameters
For an ellipse, the endpoints of the conjugate diameters can be determined using the semi-major and semi-minor axes. The endpoints of the conjugate diameters can be expressed as:
- \(P(a, 0)\) and \(Q(-a, 0)\) for the major axis
- \(P(0, b)\) and \(Q(0, -b)\) for the minor axis

In our case, we can take:
\[ P(3, 0) \quad \text{and} \quad Q(-3, 0) \]
or
\[ P(0, 2) \quad \text{and} \quad Q(0, -2) \]

### Step 4: Calculate the area of triangle CPQ
Using the points \(P(3, 0)\) and \(Q(0, 2)\), we can calculate the area of triangle \(CPQ\) using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base can be the distance between points \(P\) and \(Q\) along the x-axis, which is \(3\) (from \(0\) to \(3\)), and the height is the y-coordinate of point \(Q\), which is \(2\).

Thus, the area is:
\[ \text{Area} = \frac{1}{2} \times 3 \times 2 = 3 \]

### Final Answer
The area of triangle \(CPQ\) is \(3\) square units.

---