Find the equation of the normal to curve `y^(2)=4x` at the point `(1, 2)`.

Answer

To find the equation of the normal to the curve \( y^2 = 4x \) at the point \( (1, 2) \), we can follow these steps:

### Step 1: Differentiate the curve
We start with the equation of the curve:
\[ y^2 = 4x \]
To find the slope of the tangent line at any point on the curve, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4x) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 4 \]
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

### Step 2: Find the slope of the tangent at the point (1, 2)
Now we substitute the \( y \)-coordinate of the given point \( (1, 2) \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(1, 2)} = \frac{2}{2} = 1 \]
Thus, the slope of the tangent line at the point \( (1, 2) \) is \( m_t = 1 \).

### Step 3: Find the slope of the normal
The slope of the normal line \( m_n \) is the negative reciprocal of the slope of the tangent line:
\[ m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1 \]

### Step 4: Use the point-slope form to find the equation of the normal
Now that we have the slope of the normal line and a point on the line, we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (1, 2) \) and \( m = -1 \):
\[ y - 2 = -1(x - 1) \]
Simplifying this:
\[ y - 2 = -x + 1 \]
\[ y = -x + 3 \]

### Step 5: Rearranging to standard form
We can rearrange the equation to standard form:
\[ x + y - 3 = 0 \]

Thus, the equation of the normal to the curve \( y^2 = 4x \) at the point \( (1, 2) \) is:
\[ x + y - 3 = 0 \]