Find the slope of normal to the curve if equation of the curve is `y^2=4x` at ( 4, 5)

Answer

To find the slope of the normal to the curve given by the equation \( y^2 = 4x \) at the point \( (4, 5) \), we will follow these steps:

### Step 1: Differentiate the curve equation
We start with the equation of the curve:
\[ y^2 = 4x \]
To find the slope of the tangent line, we need to differentiate both sides with respect to \( x \).

### Step 2: Apply implicit differentiation
Differentiating \( y^2 \) gives us:
\[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]
Differentiating \( 4x \) gives us:
\[ \frac{d}{dx}(4x) = 4 \]
So, we have:
\[ 2y \frac{dy}{dx} = 4 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Now, we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

### Step 4: Substitute the point (4, 5)
Next, we substitute the \( y \)-coordinate of the given point \( (4, 5) \) into the derivative to find the slope of the tangent line:
\[ \frac{dy}{dx} \bigg|_{(4, 5)} = \frac{2}{5} \]
Let’s denote this slope as \( m_1 \):
\[ m_1 = \frac{2}{5} \]

### Step 5: Find the slope of the normal
The slope of the normal line \( m_2 \) is related to the slope of the tangent line by the fact that they are perpendicular:
\[ m_1 \cdot m_2 = -1 \]
Substituting \( m_1 \):
\[ \frac{2}{5} \cdot m_2 = -1 \]

### Step 6: Solve for \( m_2 \)
Now, we can solve for \( m_2 \):
\[ m_2 = -\frac{5}{2} \]

### Conclusion
Thus, the slope of the normal to the curve at the point \( (4, 5) \) is:
\[ \boxed{-\frac{5}{2}} \]