Let `'theta'` be the angle in radians between the curves `(x ^(2))/(36) + (y^(2))/(4) =1 and x ^(2) +y^(2) =12.` If `theta = tan ^(-1) ((a )/(sqrt3)),` Find the value of a.

Answer

To solve the problem, we need to find the angle \( \theta \) between the two curves given by the equations:

1. \( \frac{x^2}{36} + \frac{y^2}{4} = 1 \) (Ellipse)
2. \( x^2 + y^2 = 12 \) (Circle)

We will follow these steps:

### Step 1: Find Points of Intersection
To find the points of intersection of the two curves, we can substitute \( y^2 \) from the circle's equation into the ellipse's equation.

From the circle's equation:
\[ y^2 = 12 - x^2 \]

Substituting into the ellipse's equation:
\[ \frac{x^2}{36} + \frac{12 - x^2}{4} = 1 \]

### Step 2: Simplify the Equation
Multiply through by 36 to eliminate the denominators:
\[ x^2 + 9(12 - x^2) = 36 \]
\[ x^2 + 108 - 9x^2 = 36 \]
\[ -8x^2 + 108 = 36 \]
\[ -8x^2 = 36 - 108 \]
\[ -8x^2 = -72 \]
\[ x^2 = 9 \]
Thus, \( x = \pm 3 \).

### Step 3: Find Corresponding y-values
Using \( x^2 = 9 \):
\[ y^2 = 12 - 9 = 3 \quad \Rightarrow \quad y = \pm \sqrt{3} \]

The points of intersection are:
\[ (3, \sqrt{3}), (3, -\sqrt{3}), (-3, \sqrt{3}), (-3, -\sqrt{3}) \]

### Step 4: Find the Slopes of the Tangents
**For the Circle:**
The equation of the circle is \( x^2 + y^2 = 12 \).
Differentiating implicitly:
\[ 2x + 2y \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y} \]
At point \( (3, \sqrt{3}) \):
\[ \frac{dy}{dx} = -\frac{3}{\sqrt{3}} = -\sqrt{3} \]

**For the Ellipse:**
The equation of the ellipse is \( \frac{x^2}{36} + \frac{y^2}{4} = 1 \).
Differentiating implicitly:
\[ \frac{2x}{36} + \frac{2y}{4} \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x/18}{y/2} = -\frac{x}{9y} \]
At point \( (3, \sqrt{3}) \):
\[ \frac{dy}{dx} = -\frac{3}{9\sqrt{3}} = -\frac{1}{3\sqrt{3}} \]

### Step 5: Find the Angle Between the Tangents
Using the formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Here, \( m_1 = -\sqrt{3} \) and \( m_2 = -\frac{1}{3\sqrt{3}} \):
\[ \tan \theta = \left| \frac{-\sqrt{3} + \frac{1}{3\sqrt{3}}}{1 + \left(-\sqrt{3}\right)\left(-\frac{1}{3\sqrt{3}}\right)} \right| \]
Calculating the numerator:
\[ -\sqrt{3} + \frac{1}{3\sqrt{3}} = -\frac{3\sqrt{3}}{3} + \frac{1}{3\sqrt{3}} = \frac{-3\sqrt{3} + 1}{3\sqrt{3}} \]
Calculating the denominator:
\[ 1 + \frac{1}{3} = \frac{4}{3} \]
Thus,
\[ \tan \theta = \left| \frac{\frac{-3\sqrt{3} + 1}{3\sqrt{3}}}{\frac{4}{3}} \right| = \left| \frac{-3\sqrt{3} + 1}{4\sqrt{3}} \right| \]

### Step 6: Set Equal to Given Expression
We know \( \theta = \tan^{-1} \left( \frac{a}{\sqrt{3}} \right) \), so we equate:
\[ \frac{a}{\sqrt{3}} = \frac{-3\sqrt{3} + 1}{4\sqrt{3}} \]
Multiplying both sides by \( \sqrt{3} \):
\[ a = \frac{-3\cdot 3 + 1}{4} = \frac{-9 + 1}{4} = \frac{-8}{4} = -2 \]

### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{-2} \]