The point `(3tan (theta +60^(@)),2 tan(theta +30^(@)))` lies on the conic, then its centre is `(theta` is the parameter)

Options

A. `(-3sqrt(3),2sqrt(3))`
B. `(3sqrt(3),-2sqrt(3))`
C. `(-3sqrt(3),-2sqrt(3))`
D. (0,0)

Explanation

To find the center of the conic given the point \((3 \tan(\theta + 60^\circ), 2 \tan(\theta + 30^\circ))\), we can follow these steps:

### Step 1: Define the Point
Let:
\[ x = 3 \tan(\theta + 60^\circ) \]
\[ y = 2 \tan(\theta + 30^\circ) \]

### Step 2: Express Tangents in Terms of x and y
From the definitions above, we can express \(\tan(\theta + 60^\circ)\) and \(\tan(\theta + 30^\circ)\) as:
\[ \tan(\theta + 60^\circ) = \frac{x}{3} \]
\[ \tan(\theta + 30^\circ) = \frac{y}{2} \]

### Step 3: Use the Tangent Subtraction Formula
We know that:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Let \(A = \theta + 60^\circ\) and \(B = \theta + 30^\circ\). Then:
\[ \tan((\theta + 60^\circ) - (\theta + 30^\circ)) = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Using the formula:
\[ \tan(30^\circ) = \frac{\tan(\theta + 60^\circ) - \tan(\theta + 30^\circ)}{1 + \tan(\theta + 60^\circ) \tan(\theta + 30^\circ)} \]
Substituting the expressions for \(\tan(\theta + 60^\circ)\) and \(\tan(\theta + 30^\circ)\):
\[ \frac{1}{\sqrt{3}} = \frac{\frac{x}{3} - \frac{y}{2}}{1 + \frac{x}{3} \cdot \frac{y}{2}} \]

### Step 4: Cross Multiply
Cross multiplying gives:
\[ 1 + \frac{xy}{6} = \sqrt{3} \left(\frac{x}{3} - \frac{y}{2}\right) \]
This simplifies to:
\[ 1 + \frac{xy}{6} = \frac{\sqrt{3}x}{3} - \frac{\sqrt{3}y}{2} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{xy}{6} + \frac{\sqrt{3}y}{2} - \frac{\sqrt{3}x}{3} + 1 = 0 \]
Multiplying through by 6 to eliminate the fraction:
\[ xy + 3\sqrt{3}y - 2\sqrt{3}x + 6 = 0 \]

### Step 6: Identify the Center
This is a conic equation in the form:
\[ Ax + By + C = 0 \]
To find the center, we can rewrite it as:
\[ x - \frac{3\sqrt{3}}{2} + y - 2\sqrt{3} + 24 = 0 \]
The center \((h, k)\) of the conic can be found by identifying:
\[ h = \frac{3\sqrt{3}}{2}, \quad k = 2\sqrt{3} \]

### Final Answer
Thus, the center of the conic is:
\[ \left(-\frac{3\sqrt{3}}{2}, 2\sqrt{3}\right) \]