Find the equation of the normal line to the parabola  
y
=
x
2
−
8
x
+
7
 , that is also parallel to  
−
4
x
−
5
y
=
−
1
.
 

We need to find the equation of the normal line to the parabola \( y = x^2 - 8x + 7 \), which is parallel to the line \( -4x - 5y = -1 \).

### Step 1: Rewrite the given line equation in slope-intercept form
First, we need to find the slope of the given line \( -4x - 5y = -1 \). Let's rearrange this equation into slope-intercept form \( y = mx + b \).

\[
-5y = 4x - 1
\]
\[
y = -\frac{4}{5}x + \frac{1}{5}
\]

So, the slope of the line is \( m = -\frac{4}{5} \).

### Step 2: Find the slope of the normal line
Since the normal line is parallel to this line, the normal line must also have the same slope, \( m_{\text{normal}} = -\frac{4}{5} \).

### Step 3: Find the derivative of the parabola (slope of the tangent line)
The equation of the parabola is \( y = x^2 - 8x + 7 \). To find the slope of the tangent line at any point on the parabola, we need to take the derivative with respect to \( x \).

\[
\frac{dy}{dx} = 2x - 8
\]

This is the slope of the tangent line at any point \( x \) on the parabola.

### Step 4: Set up the relationship between the slopes
Since the normal line is perpendicular to the tangent line, their slopes must satisfy the condition:

\[
m_{\text{tangent}} \times m_{\text{normal}} = -1
\]

Substitute \( m_{\text{normal}} = -\frac{4}{5} \):

\[
(2x - 8) \times -\frac{4}{5} = -1
\]

### Step 5: Solve for \( x \)
Simplify the equation:

\[
(2x - 8) \times -\frac{4}{5} = -1
\]
\[
\frac{4(2x - 8)}{5} = 1
\]
\[
4(2x - 8) = 5
\]
\[
8x - 32 = 5
\]
\[
8x = 37
\]
\[
x = \frac{37}{8}
\]

### Step 6: Find the corresponding \( y \)-coordinate
Now that we have \( x = \frac{37}{8} \), substitute this into the equation of the parabola to find the corresponding \( y \)-coordinate.

\[
y = \left(\frac{37}{8}\right)^2 - 8\left(\frac{37}{8}\right) + 7
\]
\[
y = \frac{1369}{64} - \frac{296}{8} + 7
\]
\[
y = \frac{1369}{64} - \frac{2368}{64} + \frac{448}{64}
\]
\[
y = \frac{1369 - 2368 + 448}{64} = \frac{-551}{64}
\]

So, the point on the parabola is \( \left(\frac{37}{8}, \frac{-551}{64}\right) \).

### Step 7: Find the equation of the normal line
Now that we have the slope of the normal line \( m = -\frac{4}{5} \) and a point on the line \( \left(\frac{37}{8}, \frac{-551}{64}\right) \), we can use the point-slope form of the equation of a line:

\[
y - y_1 = m(x - x_1)
\]

Substitute \( m = -\frac{4}{5} \), \( x_1 = \frac{37}{8} \), and \( y_1 = \frac{-551}{64} \):

\[
y - \frac{-551}{64} = -\frac{4}{5}\left(x - \frac{37}{8}\right)
\]

### Step 8: Simplify the equation
Let's multiply both sides to get rid of the fractions and simplify:

\[
y + \frac{551}{64} = -\frac{4}{5}(x - \frac{37}{8})
\]

Multiplying everything out will give the final equation of the normal line, which you can simplify further if necessary.

Would you like help simplifying or have questions?

---

### Related questions:
1. How do you find the tangent line to a parabola at a given point?
2. What is the difference between a normal line and a tangent line?
3. How do you determine if two lines are parallel or perpendicular?
4. What is the geometric interpretation of the derivative in calculus?
5. How do you convert between point-slope form and slope-intercept form?

### Tip:
When working with slopes of normal and tangent lines, always remember that the product of their slopes must equal \(-1\) for perpendicularity!

Find the equation of the normal line to the parabola y = x^2 - 8x + 7, that is also parallel to -4x - 5y = -1.

Learn how to find the equation of a normal line to the parabola y = x^2 - 8x + 7 that is parallel to the line -4x - 5y = -1. This solution explores key concepts in algebra and calculus, including the derivative for finding slopes, and the relationship between perpendicular lines. Suitable for students in Grades 11-12 or those studying pre-calculus and calculus.

Math Problem Statement

Find the equation of the normal line to the parabola
y

x 2 − 8 x + 7 , that is also parallel to
− 4 x − 5 y

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