If `theta` is the angle between the pair of tangents drawn to the ellipse `3x^(2)+2y^(2)=5` from the point (1, 2), then the value of `tan^(2)theta` is equal to

Answer

To find the value of \( \tan^2 \theta \) where \( \theta \) is the angle between the pair of tangents drawn to the ellipse \( 3x^2 + 2y^2 = 5 \) from the point \( (1, 2) \), we can follow these steps:

### Step 1: Write the equation of the ellipse in standard form
The given ellipse is:
\[ 3x^2 + 2y^2 = 5 \]
Dividing both sides by 5, we get:
\[ \frac{x^2}{\frac{5}{3}} + \frac{y^2}{\frac{5}{2}} = 1 \]
This shows that \( a^2 = \frac{5}{3} \) and \( b^2 = \frac{5}{2} \).

### Step 2: Identify the coordinates of the point
The point from which the tangents are drawn is \( (x_1, y_1) = (1, 2) \).

### Step 3: Use the formula for the pair of tangents from a point to an ellipse
The equation for the pair of tangents drawn from a point \( (x_1, y_1) \) to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \( a^2 \) and \( b^2 \):
\[ \frac{xx_1}{\frac{5}{3}} + \frac{yy_1}{\frac{5}{2}} = 1 \]
This simplifies to:
\[ \frac{3x}{5} + \frac{2y}{5} = 1 \]
Multiplying through by 5 gives:
\[ 3x + 2y = 5 \]

### Step 4: Find the coefficients for the angle between the tangents
The general form of the equation of the pair of tangents is:
\[ Ax^2 + By^2 + 2Hxy + 2Fx + 2Gy + C = 0 \]
For our case, we have:
- \( A = 3 \)
- \( B = 2 \)
- \( H = 0 \)
- \( F = -5 \)
- \( G = 0 \)
- \( C = 0 \)

### Step 5: Calculate \( \tan^2 \theta \)
Using the formula:
\[ \tan^2 \theta = \frac{2\sqrt{H^2 - AB}}{A + B} \]
Substituting the values:
\[ \tan^2 \theta = \frac{2\sqrt{0^2 - (3)(2)}}{3 + 2} = \frac{2\sqrt{-6}}{5} \]
Since \( \tan^2 \theta \) cannot be negative, we need to use the discriminant method for the quadratic formed by the tangents.

### Step 6: Find the discriminant
The discriminant \( D \) of the quadratic equation formed by the tangents is given by:
\[ D = (2F)^2 - 4AB \]
Substituting the values:
\[ D = (-10)^2 - 4(3)(2) = 100 - 24 = 76 \]
Thus, the angle \( \theta \) is given by:
\[ \tan^2 \theta = \frac{D}{(A + B)^2} = \frac{76}{(5)^2} = \frac{76}{25} \]

### Final Answer
Thus, the value of \( \tan^2 \theta \) is:
\[ \boxed{\frac{76}{25}} \]