If the line `y=xtantheta ` cut the curve `x^3+xy^2+2x^2+2y^2+3x+1=0` at the points A,B and C. if OA, OB and OC are in HP then `tan theta ` is equal to

Options

A. `pm1`
B. 0
C. 2
D. `-2`

Answer

To solve the problem, we need to find the value of \( \tan \theta \) given that the line \( y = x \tan \theta \) intersects the curve \( x^3 + xy^2 + 2x^2 + 2y^2 + 3x + 1 = 0 \) at points A, B, and C, and that the distances \( OA, OB, OC \) are in Harmonic Progression (HP).

### Step-by-Step Solution:

1. **Substitute the Line Equation into the Curve:**
We start by substituting \( y = x \tan \theta \) into the curve equation:
\[ x^3 + x(x \tan \theta)^2 + 2x^2 + 2(x \tan \theta)^2 + 3x + 1 = 0 \]
This simplifies to:
\[ x^3 + x^3 \tan^2 \theta + 2x^2 + 2x^2 \tan^2 \theta + 3x + 1 = 0 \]
Combining like terms gives:
\[ x^3(1 + \tan^2 \theta) + x^2(2 + 2\tan^2 \theta) + 3x + 1 = 0 \]

2. **Identify the Cubic Equation:**
The equation can be rewritten as:
\[ (1 + \tan^2 \theta)x^3 + (2 + 2\tan^2 \theta)x^2 + 3x + 1 = 0 \]
Let \( k = 1 + \tan^2 \theta \) and \( m = 2 + 2\tan^2 \theta \). The cubic equation becomes:
\[ kx^3 + mx^2 + 3x + 1 = 0 \]

3. **Roots of the Cubic Equation:**
Let the roots of this cubic equation be \( x_1, x_2, x_3 \) corresponding to points A, B, and C. We know that \( OA, OB, OC \) are in HP, which implies:
\[ \frac{1}{OA}, \frac{1}{OB}, \frac{1}{OC} \text{ are in AP} \]

4. **Calculate Distances from Origin:**
The distances from the origin to points A, B, and C are given by:
\[ OA = \sqrt{x_1^2 + (x_1 \tan \theta)^2} = x_1 \sec \theta \]
\[ OB = \sqrt{x_2^2 + (x_2 \tan \theta)^2} = x_2 \sec \theta \]
\[ OC = \sqrt{x_3^2 + (x_3 \tan \theta)^2} = x_3 \sec \theta \]

5. **Use the Harmonic Progression Condition:**
Since \( OA, OB, OC \) are in HP, we have:
\[ \frac{2}{OB} = \frac{1}{OA} + \frac{1}{OC} \]
Substituting the distances:
\[ \frac{2}{x_2 \sec \theta} = \frac{1}{x_1 \sec \theta} + \frac{1}{x_3 \sec \theta} \]
This simplifies to:
\[ 2 = \frac{x_2}{x_1} + \frac{x_2}{x_3} \]
Rearranging gives:
\[ 2x_1 x_3 = x_1 + x_3 \]

6. **Relate the Roots to Coefficients:**
From Vieta's formulas, we know:
- \( x_1 + x_2 + x_3 = -\frac{m}{k} \)
- \( x_1 x_2 + x_2 x_3 + x_1 x_3 = \frac{3}{k} \)
- \( x_1 x_2 x_3 = -\frac{1}{k} \)

7. **Substituting Values:**
We can substitute \( x_2 = -1 \) into the equations derived from Vieta's formulas and solve for \( \tan^2 \theta \):
\[ x_1 + (-1) + x_3 = -\frac{m}{k} \]
\[ x_1(-1) + (-1)x_3 + x_1 x_3 = \frac{3}{k} \]
\[ x_1(-1)(x_3) = -\frac{1}{k} \]

8. **Solving for \( \tan^2 \theta \):**
After substituting and simplifying, we find:
\[ \sec^2 \theta = 2 \implies \tan^2 \theta = 1 \]
Thus, \( \tan \theta = \pm 1 \).

### Final Answer:
\[ \tan \theta = \pm 1 \]