y=x is tangent to the parabola `y=ax^(2)+c`.
If (1,1) is the point of contact, then a is (a) `1/4` (b) `1/3` (c) `1/2` (d) `1/6`

Options

A. `1//4`
B. `1//3`
C. `1//2`
D. `1//6`

Explanation

To find the value of \( a \) such that the line \( y = x \) is tangent to the parabola \( y = ax^2 + c \) at the point \( (1, 1) \), we can follow these steps:

### Step 1: Set up the equations
We know the equations of the line and the parabola:
- Line: \( y = x \)
- Parabola: \( y = ax^2 + c \)

### Step 2: Substitute the point of contact into the parabola's equation
Since \( (1, 1) \) is the point of contact, we substitute \( x = 1 \) and \( y = 1 \) into the parabola's equation:
\[ 1 = a(1)^2 + c \]
This simplifies to:
\[ 1 = a + c \quad \text{(Equation 1)} \]

### Step 3: Find the derivative of the parabola
To find the slope of the tangent to the parabola at any point, we take the derivative of \( y = ax^2 + c \):
\[ \frac{dy}{dx} = 2ax \]

### Step 4: Evaluate the derivative at the point of contact
At the point \( (1, 1) \), we substitute \( x = 1 \) into the derivative:
\[ \frac{dy}{dx} = 2a(1) = 2a \]

### Step 5: Set the slope equal to the slope of the line
Since the slope of the line \( y = x \) is \( 1 \), we set the derivative equal to \( 1 \):
\[ 2a = 1 \]

### Step 6: Solve for \( a \)
Now, we solve for \( a \):
\[ a = \frac{1}{2} \]

### Step 7: Verify the value of \( c \)
Using Equation 1, we can find \( c \):
\[ 1 = \frac{1}{2} + c \implies c = 1 - \frac{1}{2} = \frac{1}{2} \]

### Conclusion
Thus, the value of \( a \) is \( \frac{1}{2} \).

### Final Answer
The correct option is (c) \( \frac{1}{2} \).

---