The set of points on the axis of the parabola `y^2-4x-2y+5=0` find the slope of normal to the curve at (0,0)

Answer

To find the slope of the normal to the curve defined by the equation \( y^2 - 4x - 2y + 5 = 0 \) at the point \( (0,0) \), we will follow these steps:

### Step 1: Rewrite the equation of the parabola
We start with the equation of the parabola:
\[ y^2 - 4x - 2y + 5 = 0 \]
We can rearrange it to express \( y \) in terms of \( x \):
\[ y^2 - 2y + 5 = 4x \]

### Step 2: Differentiate the equation implicitly
Next, we differentiate both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(2y) + \frac{d}{dx}(5) = \frac{d}{dx}(4x) \]
Using the chain rule, we get:
\[ 2y \frac{dy}{dx} - 2 \frac{dy}{dx} = 4 \]

### Step 3: Factor out \( \frac{dy}{dx} \)
Now, we can factor out \( \frac{dy}{dx} \):
\[ (2y - 2) \frac{dy}{dx} = 4 \]

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

### Step 5: Evaluate \( \frac{dy}{dx} \) at the point \( (0,0) \)
We substitute \( y = 0 \) into the derivative:
\[ \frac{dy}{dx} = \frac{4}{2(0) - 2} = \frac{4}{-2} = -2 \]
Thus, the slope of the tangent line at the point \( (0,0) \) is \( -2 \).

### Step 6: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent. If the slope of the tangent is \( m_1 = -2 \), then the slope of the normal \( m_2 \) is given by:
\[ m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1} = -\frac{1}{-2} = \frac{1}{2} \]

### Conclusion
The slope of the normal to the curve at the point \( (0, 0) \) is:
\[ \boxed{\frac{1}{2}} \]