If `sqrt(3)bx +ay=2ab` is a tangent to the ellipse then ecentric angle `theta` is

Options

A. `(pi)/(4)`
B. `(pi)/(6)`
C. `(pi)/(2)`
D. `(pi)/(5)`

Answer

To solve the problem, we need to determine the eccentric angle \( \theta \) given that the line \( \sqrt{3}bx + ay = 2ab \) is a tangent to the ellipse. We will follow these steps:

### Step 1: Write the equation of the tangent line
The given equation of the tangent line is:
\[ \sqrt{3}bx + ay = 2ab \]

### Step 2: Rearranging the equation
We can rearrange this equation to the standard form of a line:
\[ \sqrt{3}bx + ay - 2ab = 0 \]

### Step 3: Compare with the standard tangent equation
The standard equation of the tangent to the ellipse at the point corresponding to the eccentric angle \( \theta \) is given by:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]
We can rewrite this as:
\[ x \cos \theta + y \sin \theta = a \]

### Step 4: Normalize the tangent line equation
To compare the two equations, we will divide the entire equation of the tangent line by \( 2ab \):
\[ \frac{\sqrt{3}bx}{2ab} + \frac{ay}{2ab} = 1 \]
This simplifies to:
\[ \frac{\sqrt{3}x}{2a} + \frac{y}{2b} = 1 \]

### Step 5: Set up the comparison
Now we have two equations:
1. \( \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \) (Equation of the tangent to the ellipse)
2. \( \frac{\sqrt{3}x}{2a} + \frac{y}{2b} = 1 \) (Normalized tangent line)

From these, we can compare coefficients:
- Comparing the coefficients of \( x \):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
- Comparing the coefficients of \( y \):
\[ \sin \theta = \frac{1}{2} \]

### Step 6: Find the eccentric angle \( \theta \)
From the trigonometric values, we know:
\[ \cos \theta = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin \theta = \frac{1}{2} \]
This corresponds to:
\[ \theta = 30^\circ = \frac{\pi}{6} \]

### Final Answer
Thus, the eccentric angle \( \theta \) is:
\[ \theta = \frac{\pi}{6} \]

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