The straight line `x + y = a` touches the parabola `y=x-x^2` If a =

Options

A. 0
B. 1
C. -1
D. none

Answer

To find the value of \( a \) for which the line \( x + y = a \) touches the parabola \( y = x - x^2 \), we can follow these steps:

### Step 1: Rewrite the line equation
The line equation \( x + y = a \) can be rewritten in slope-intercept form:
\[ y = a - x \]

### Step 2: Set the equations equal to find intersection points
To find the points of intersection between the line and the parabola, we set the equations equal to each other:
\[ a - x = x - x^2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ x^2 - 2x + a = 0 \]

### Step 4: Use the condition for tangency
For the line to touch the parabola, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The discriminant \( D \) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
In our case, \( a = 1 \), \( b = -2 \), and \( c = a \). Thus, we have:
\[ D = (-2)^2 - 4(1)(a) = 4 - 4a \]

### Step 5: Set the discriminant equal to zero
Setting the discriminant equal to zero for tangency:
\[ 4 - 4a = 0 \]

### Step 6: Solve for \( a \)
Solving for \( a \):
\[ 4 = 4a \implies a = 1 \]

### Conclusion
The value of \( a \) for which the line \( x + y = a \) touches the parabola \( y = x - x^2 \) is:
\[ \boxed{1} \]