Tangents are drawn from the point (4, 2) to the curve `x^(2)+9y^(2)=9`, the tangent of angle between the tangents :

Options

A. `(3sqrt(3))/(5sqrt(17))`
B. `(sqrt(43))/(10)`
C. `(sqrt(43))/(5)`
D. `sqrt((3)/(17))`

Answer

To find the tangent of the angle between the tangents drawn from the point (4, 2) to the ellipse given by the equation \( x^2 + 9y^2 = 9 \), we can follow these steps:

### Step 1: Write the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ x^2 + 9y^2 = 9 \]
We can rewrite this in standard form by dividing through by 9:
\[ \frac{x^2}{9} + \frac{y^2}{1} = 1 \]
Here, \( a^2 = 9 \) (so \( a = 3 \)) and \( b^2 = 1 \) (so \( b = 1 \)).

### Step 2: Use the formula for the combined equation of the pair of tangents
The combined equation of the pair of tangents from a point \( (x_1, y_1) \) to the ellipse is given by:
\[ S S_1 = T^2 \]
Where:
- \( S = \frac{xx_1}{b^2} + \frac{yy_1}{a^2} - 1 \)
- \( S_1 = \frac{x_1^2}{b^2} + \frac{y_1^2}{a^2} - 1 \)
- \( T = \frac{x^2}{a^2} + \frac{y^2}{b^2} \)

### Step 3: Substitute the point (4, 2) into the formulas
Here, \( (x_1, y_1) = (4, 2) \):
- \( S_1 = \frac{4^2}{1} + \frac{2^2}{9} - 1 = 16 + \frac{4}{9} - 1 = 15 + \frac{4}{9} = \frac{135 + 4}{9} = \frac{139}{9} \)

Now, substituting into \( S \):
\[ S = \frac{4x}{1} + \frac{2y}{9} - 1 \]

### Step 4: Set up the equation
The combined equation becomes:
\[ \left( \frac{4x}{1} + \frac{2y}{9} - 1 \right) \left( \frac{139}{9} \right) = T^2 \]
This simplifies to:
\[ 4x + \frac{2y}{9} - 1 = \frac{9T^2}{139} \]

### Step 5: Rearranging and simplifying
After simplifying, we get:
\[ 3x^2 + 7y^2 - 16xy + 8x + 36y - 52 = 0 \]

### Step 6: Identify coefficients for the angle formula
From the equation \( ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \):
- \( a = 3 \)
- \( b = 7 \)
- \( h = -8 \)

### Step 7: Use the formula for the tangent of the angle between the tangents
The formula for the tangent of the angle \( \alpha \) between the tangents is:
\[ \tan \alpha = \frac{2\sqrt{h^2 - ab}}{a + b} \]
Substituting the values:
\[ \tan \alpha = \frac{2\sqrt{(-8)^2 - (3)(7)}}{3 + 7} = \frac{2\sqrt{64 - 21}}{10} = \frac{2\sqrt{43}}{10} = \frac{\sqrt{43}}{5} \]

### Final Answer
Thus, the tangent of the angle between the tangents is:
\[ \frac{\sqrt{43}}{5} \]