If the line y = mx + a meets the parabola `x^(2)=4ay` in two points whose
abscissa are `x_(1)` and `x_(2)` then `x_(1)+x_(2)` =0 If

Options

A. `m= -1`
B. `m=1`
C. m=0
D. `m=-(1)/(2)`

Answer

To solve the problem, we need to find the condition under which the line \( y = mx + a \) intersects the parabola \( x^2 = 4ay \) at two points whose abscissas (x-coordinates) sum to zero.

### Step-by-step Solution:

1. **Substitute the line equation into the parabola equation:**
We start by substituting \( y \) from the line equation into the parabola equation.
\[ x^2 = 4a(mx + a) \]

2. **Rearrange the equation:**
Expanding the right-hand side gives:
\[ x^2 = 4amx + 4a^2 \]
Rearranging this, we get:
\[ x^2 - 4amx - 4a^2 = 0 \]

3. **Identify the quadratic equation:**
The equation we have is a standard quadratic form \( Ax^2 + Bx + C = 0 \), where:
- \( A = 1 \)
- \( B = -4am \)
- \( C = -4a^2 \)

4. **Use the sum of the roots formula:**
The sum of the roots \( x_1 + x_2 \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ x_1 + x_2 = -\frac{B}{A} \]
Substituting our values:
\[ x_1 + x_2 = -\frac{-4am}{1} = 4am \]

5. **Set the sum of the roots to zero:**
According to the problem, we know that:
\[ x_1 + x_2 = 0 \]
Therefore, we can set the equation:
\[ 4am = 0 \]

6. **Solve for \( m \):**
Since \( 4am = 0 \), we have two cases:
- \( 4a = 0 \) which implies \( a = 0 \) (but this would not yield a parabola).
- \( m = 0 \) (this is the valid solution).

Thus, the only solution that satisfies the given condition is:
\[ m = 0 \]

### Conclusion:
The value of \( m \) must be \( 0 \) for the line \( y = mx + a \) to meet the parabola \( x^2 = 4ay \) at two points whose abscissas sum to zero.