Any chord of the conic `x^(2)+y^(2)+xy=1` passing through origin is bisected at a point (p, q), then `(p+q+12)` equals to :

Options

A. A) 13
B. B) 14
C. C) 11
D. D) 12

Answer

To solve the problem, we need to analyze the given conic and the properties of the chord passing through the origin.

### Step-by-Step Solution:

1. **Identify the Conic Equation**:
The equation of the conic is given as:
\[ x^2 + y^2 + xy = 1 \]

2. **Equation of the Chord**:
Since the chord passes through the origin, we can express the line in slope-intercept form as:
\[ y = mx \]
where \( m \) is the slope of the line.

3. **Substituting the Line Equation into the Conic**:
Substitute \( y = mx \) into the conic equation:
\[ x^2 + (mx)^2 + x(mx) = 1 \]
This simplifies to:
\[ x^2 + m^2x^2 + mx^2 = 1 \]
or
\[ (1 + m^2 + m)x^2 = 1 \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ x^2(1 + m + m^2) = 1 \]
This can be rewritten as:
\[ x^2 = \frac{1}{1 + m + m^2} \]

5. **Finding the Roots**:
Let the roots of this equation be \( x_1 \) and \( x_2 \). From the quadratic equation, we know that the sum of the roots \( x_1 + x_2 \) is given by:
\[ x_1 + x_2 = 0 \]
(since there is no \( x \) term in the equation).

6. **Finding the Midpoint**:
The midpoint (or the bisecting point) of the chord, which is given as \( (p, q) \), can be calculated as:
\[ p = \frac{x_1 + x_2}{2} = \frac{0}{2} = 0 \]
For the y-coordinates:
\[ y_1 = mx_1 \quad \text{and} \quad y_2 = mx_2 \]
Thus,
\[ y_1 + y_2 = mx_1 + mx_2 = m(x_1 + x_2) = m(0) = 0 \]
Therefore,
\[ q = \frac{y_1 + y_2}{2} = \frac{0}{2} = 0 \]

7. **Calculating \( p + q + 12 \)**:
Now, we can calculate:
\[ p + q + 12 = 0 + 0 + 12 = 12 \]

### Final Answer:
Thus, the value of \( p + q + 12 \) is:
\[ \boxed{12} \]