Find the slope of normal to the curve if equation of the curve is `y^2=4ax` at (0 , 0)

Answer

To find the slope of the normal to the curve given by the equation \( y^2 = 4ax \) at the point \( (0, 0) \), we can follow these steps:

### Step 1: Differentiate the equation of the curve
The equation of the curve is given by:
\[ y^2 = 4ax \]
We will differentiate both sides with respect to \( x \).

### Step 2: Apply implicit differentiation
Differentiating \( y^2 \) with respect to \( x \) gives:
\[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]
Differentiating \( 4ax \) with respect to \( x \) gives:
\[ \frac{d}{dx}(4ax) = 4a \]
Setting these equal gives:
\[ 2y \frac{dy}{dx} = 4a \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at the point (0, 0)
Now we need to find the slope of the tangent line at the point \( (0, 0) \). Substituting \( y = 0 \) into the derivative:
\[ \frac{dy}{dx} = \frac{2a}{0} \]
This expression is undefined, indicating that the slope of the tangent at this point is vertical.

### Step 5: Determine the slope of the normal
The slope of the normal line is perpendicular to the slope of the tangent line. Since the tangent line is vertical, the slope of the normal line will be horizontal. Therefore, the slope of the normal line is:
\[ m_2 = 0 \]

### Final Answer
The slope of the normal to the curve \( y^2 = 4ax \) at the point \( (0, 0) \) is:
\[ \text{slope of normal} = 0 \]