If equation of the curve is `y^2=4x` then find the slope of normal to the curve

Answer

To find the slope of the normal to the curve given by the equation \( y^2 = 4x \), 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 to 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, we get:
\[ 2y \frac{dy}{dx} = 4 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]
This expression represents the slope of the tangent line at any point on the curve.

### Step 3: Find the slope of the normal
The slope of the normal line is related to the slope of the tangent line. If the slope of the tangent is \( m_1 \), then the slope of the normal \( m_2 \) can be found using the relationship:
\[ m_1 \cdot m_2 = -1 \]
Substituting \( m_1 = \frac{2}{y} \):
\[ \frac{2}{y} \cdot m_2 = -1 \]
Now, we can solve for \( m_2 \):
\[ m_2 = -\frac{y}{2} \]

### Final Result
Thus, the slope of the normal to the curve \( y^2 = 4x \) is:
\[ m_2 = -\frac{y}{2} \]