Find the equation of the normal to the curve `y=e^(2x)+x^(2)" at "x=0`. Also find the distance from origin to the line.

Answer

To find the equation of the normal to the curve \( y = e^{2x} + x^2 \) at \( x = 0 \) and the distance from the origin to the line, we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( y = e^{2x} + x^2 \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(x^2) \]

Using the chain rule for \( e^{2x} \) and the power rule for \( x^2 \):

\[ \frac{dy}{dx} = 2e^{2x} + 2x \]

### Step 2: Evaluate the derivative at \( x = 0 \)
Next, we substitute \( x = 0 \) into the derivative to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=0} = 2e^{2 \cdot 0} + 2 \cdot 0 = 2 \cdot 1 + 0 = 2 \]

### Step 3: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

\[ \text{slope of normal} = -\frac{1}{2} \]

### Step 4: Find the coordinates of the point on the curve at \( x = 0 \)
Now, we need to find the corresponding \( y \) value when \( x = 0 \):

\[ y = e^{2 \cdot 0} + 0^2 = 1 + 0 = 1 \]

So, the point on the curve is \( (0, 1) \).

### Step 5: Write the equation of the normal line
Using the point-slope form of the equation of a line, we can write the equation of the normal line:

\[ y - y_1 = m(x - x_1) \]

Substituting \( (x_1, y_1) = (0, 1) \) and \( m = -\frac{1}{2} \):

\[ y - 1 = -\frac{1}{2}(x - 0) \]

This simplifies to:

\[ y - 1 = -\frac{1}{2}x \]

Rearranging gives:

\[ \frac{1}{2}x + y - 1 = 0 \quad \text{or} \quad x + 2y - 2 = 0 \]

### Step 6: Find the distance from the origin to the line
The distance \( D \) from a point \( (x_0, y_0) \) to the line \( ax + by + c = 0 \) is given by:

\[ D = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]

For our line \( x + 2y - 2 = 0 \), we have \( a = 1 \), \( b = 2 \), and \( c = -2 \). The origin is \( (0, 0) \).

Substituting into the distance formula:

\[ D = \frac{|1 \cdot 0 + 2 \cdot 0 - 2|}{\sqrt{1^2 + 2^2}} = \frac{|-2|}{\sqrt{1 + 4}} = \frac{2}{\sqrt{5}} \]

### Final Answers
1. **Equation of the normal**: \( x + 2y - 2 = 0 \)
2. **Distance from the origin to the line**: \( \frac{2}{\sqrt{5}} \)