The equation of a normal to the parabola `y=x^(2)-6x+6` which is perpendicular to the line joining the origin to the vertex of the parabola is

Options

A. `4x-4y-11=0`
B. `4x-4y+1=0`
C. `4x-4y-21=0`
D. `4x-4y+21=0`

Answer

To solve the problem, we need to find the equation of a normal to the parabola \( y = x^2 - 6x + 6 \) that is perpendicular to the line joining the origin to the vertex of the parabola. Let's break down the solution step by step.

### Step 1: Find the vertex of the parabola
The given parabola is in the form \( y = ax^2 + bx + c \). We can find the vertex using the formula:
\[ x = -\frac{b}{2a} \]
For our parabola:
- \( a = 1 \)
- \( b = -6 \)

Calculating the x-coordinate of the vertex:
\[ x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3 \]

Now, substitute \( x = 3 \) back into the equation to find the y-coordinate:
\[ y = (3)^2 - 6(3) + 6 = 9 - 18 + 6 = -3 \]

Thus, the vertex of the parabola is \( (3, -3) \).

### Step 2: Find the slope of the line joining the origin to the vertex
The slope \( m \) of the line joining the origin \( (0, 0) \) and the vertex \( (3, -3) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 0}{3 - 0} = \frac{-3}{3} = -1 \]

### Step 3: Determine the slope of the normal
Since the normal is perpendicular to the line joining the origin to the vertex, the slope of the normal \( m_n \) is the negative reciprocal of \( -1 \):
\[ m_n = -\frac{1}{-1} = 1 \]

### Step 4: Find the equation of the normal
The general equation of the normal to a parabola at point \( (x_1, y_1) \) is given by:
\[ y - y_1 = m_n(x - x_1) \]
We need to find a point on the parabola where the slope of the tangent is \( -1 \) (since the normal's slope is \( 1 \)).

### Step 5: Find the points on the parabola where the slope of the tangent is \( -1 \)
The derivative of the parabola gives us the slope of the tangent:
\[ \frac{dy}{dx} = 2x - 6 \]
Setting the derivative equal to \( -1 \):
\[ 2x - 6 = -1 \implies 2x = 5 \implies x = \frac{5}{2} \]

Now, substitute \( x = \frac{5}{2} \) back into the parabola to find \( y \):
\[ y = \left(\frac{5}{2}\right)^2 - 6\left(\frac{5}{2}\right) + 6 = \frac{25}{4} - 15 + 6 = \frac{25}{4} - \frac{60}{4} + \frac{24}{4} = \frac{-11}{4} \]

Thus, the point on the parabola is \( \left(\frac{5}{2}, -\frac{11}{4}\right) \).

### Step 6: Write the equation of the normal
Using the point \( \left(\frac{5}{2}, -\frac{11}{4}\right) \) and the slope \( 1 \):
\[ y - \left(-\frac{11}{4}\right) = 1\left(x - \frac{5}{2}\right) \]
This simplifies to:
\[ y + \frac{11}{4} = x - \frac{5}{2} \]
Rearranging gives:
\[ y = x - \frac{5}{2} - \frac{11}{4} \]
Finding a common denominator (4):
\[ y = x - \frac{10}{4} - \frac{11}{4} = x - \frac{21}{4} \]

### Step 7: Convert to standard form
Rearranging gives:
\[ y - x + \frac{21}{4} = 0 \implies 4y - 4x + 21 = 0 \implies 4x - 4y + 21 = 0 \]

### Final Answer
The equation of the normal to the parabola is:
\[ 4x - 4y - 21 = 0 \]