If the parabola `x^2=ay` makes an intercept of length `sqrt40` unit on the line `y-2x = 1` then `a` is equal to

Options

A. -1
B. -2
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( a \) for the parabola \( x^2 = ay \) that makes an intercept of length \( \sqrt{40} \) on the line \( y - 2x = 1 \).

### Step-by-Step Solution:

1. **Rewrite the line equation**:
The line equation \( y - 2x = 1 \) can be rewritten as:
\[ y = 2x + 1 \]

2. **Substitute the line equation into the parabola equation**:
Substitute \( y = 2x + 1 \) into the parabola equation \( x^2 = ay \):
\[ x^2 = a(2x + 1) \]
This simplifies to:
\[ x^2 - 2ax - a = 0 \]

3. **Identify the coefficients**:
The quadratic equation in \( x \) is:
\[ x^2 - 2ax - a = 0 \]
Here, \( A = 1 \), \( B = -2a \), and \( C = -a \).

4. **Use the quadratic formula to find the roots**:
The roots \( x_1 \) and \( x_2 \) can be found using the quadratic formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Plugging in the values:
\[ x = \frac{2a \pm \sqrt{(-2a)^2 - 4(1)(-a)}}{2(1)} = \frac{2a \pm \sqrt{4a^2 + 4a}}{2} \]
This simplifies to:
\[ x = a \pm \sqrt{a^2 + a} \]

5. **Calculate the distance between the roots**:
The distance \( x_2 - x_1 \) is:
\[ x_2 - x_1 = 2\sqrt{a^2 + a} \]
Therefore, the square of the distance is:
\[ (x_2 - x_1)^2 = 4(a^2 + a) \]

6. **Find the y-coordinates for the roots**:
Substitute \( x_1 \) and \( x_2 \) back into the line equation to find \( y_1 \) and \( y_2 \):
\[ y_1 = 2(a - \sqrt{a^2 + a}) + 1, \quad y_2 = 2(a + \sqrt{a^2 + a}) + 1 \]
The difference in \( y \) coordinates is:
\[ y_2 - y_1 = 2\sqrt{a^2 + a} \]
Therefore, the square of the difference is:
\[ (y_2 - y_1)^2 = 4(a^2 + a) \]

7. **Combine the distances**:
The total distance squared is given by:
\[ (x_2 - x_1)^2 + (y_2 - y_1)^2 = 4(a^2 + a) + 4(a^2 + a) = 8(a^2 + a) \]

8. **Set the total distance squared equal to 40**:
Since the length of the intercept is \( \sqrt{40} \), we have:
\[ 8(a^2 + a) = 40 \]
Dividing both sides by 8 gives:
\[ a^2 + a = 5 \]

9. **Rearranging the equation**:
Rearranging gives:
\[ a^2 + a - 5 = 0 \]

10. **Solve the quadratic equation**:
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 20}}{2} = \frac{-1 \pm \sqrt{21}}{2} \]

### Final Answer:
Thus, the values of \( a \) are:
\[ a = \frac{-1 + \sqrt{21}}{2} \quad \text{and} \quad a = \frac{-1 - \sqrt{21}}{2} \]