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

Options

A. (3,3)
B. (2,2)
C. (6,6)
D. (4,4)

Explanation

To find the point of contact where the line \( y = x \) is tangent to the parabola \( y = ax^2 + c \) with \( c = 2 \), we can follow these steps:

### Step 1: Set up the equations
We have the line equation:
\[ y = x \]
And the parabola equation:
\[ y = ax^2 + 2 \]

### Step 2: Find the slope of the tangent
The slope of the line \( y = x \) is 1. The derivative of the parabola \( y = ax^2 + 2 \) gives us the slope of the tangent at any point \( x \):
\[ \frac{dy}{dx} = 2ax \]
Setting this equal to the slope of the line, we have:
\[ 2ax = 1 \]
From this, we can solve for \( x \):
\[ x = \frac{1}{2a} \]

### Step 3: Substitute \( x \) into the parabola equation
Since the point of contact must satisfy both equations, we substitute \( x = \frac{1}{2a} \) into the parabola equation:
\[ y = a\left(\frac{1}{2a}\right)^2 + 2 \]
Calculating this gives:
\[ y = a\left(\frac{1}{4a^2}\right) + 2 = \frac{1}{4a} + 2 \]

### Step 4: Set the equations equal
Since \( y = x \) at the point of contact, we have:
\[ \frac{1}{4a} + 2 = \frac{1}{2a} \]

### Step 5: Solve for \( a \)
Rearranging gives:
\[ \frac{1}{4a} + 2 - \frac{1}{2a} = 0 \]
Finding a common denominator (which is \( 4a \)):
\[ \frac{1 + 8a - 2}{4a} = 0 \]
This simplifies to:
\[ \frac{8a - 1}{4a} = 0 \]
Thus,
\[ 8a - 1 = 0 \implies a = \frac{1}{8} \]

### Step 6: Find \( x \)
Now substituting \( a \) back into the equation for \( x \):
\[ x = \frac{1}{2a} = \frac{1}{2 \cdot \frac{1}{8}} = 4 \]

### Step 7: Find \( y \)
Since \( y = x \), we have:
\[ y = 4 \]

### Conclusion
Thus, the point of contact is:
\[ (4, 4) \]

### Final Answer
The correct option is (d) \( (4, 4) \).

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