Find the slope of the normal to the curve `y^(2)=4 ax ` at `((a)/(m^(2)),(2a)/(m))`

Answer

To find the slope of the normal to the curve \( y^2 = 4ax \) at the point \( \left( \frac{a}{m^2}, \frac{2a}{m} \right) \), we can follow these steps:

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

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

### Step 3: Substitute the point into \(\frac{dy}{dx}\)
Next, we substitute the point \( \left( \frac{a}{m^2}, \frac{2a}{m} \right) \) into the derivative:
\[ y = \frac{2a}{m} \]
Thus,
\[ \frac{dy}{dx} = \frac{2a}{\frac{2a}{m}} = m \]

### Step 4: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent. If \( m_2 \) is the slope of the tangent, then the slope of the normal \( m_1 \) is given by:
\[ m_1 = -\frac{1}{m_2} \]
Substituting \( m_2 = m \):
\[ m_1 = -\frac{1}{m} \]

### Final Result
Thus, the slope of the normal to the curve at the given point is:
\[ \text{slope of the normal} = -\frac{1}{m} \]

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