Find the equations of the normal to the curve `y=4x^(3)-3x+5`, which are perpendicular to the line : `9x-y+5=0`.

Answer

To find the equations of the normal to the curve \( y = 4x^3 - 3x + 5 \) that are perpendicular to the line \( 9x - y + 5 = 0 \), we will follow these steps:

### Step 1: Find the slope of the tangent to the curve
First, we differentiate the function \( y \) to find the slope of the tangent.

\[ \frac{dy}{dx} = \frac{d}{dx}(4x^3 - 3x + 5) = 12x^2 - 3 \]

### Step 2: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent. Therefore, if the slope of the tangent is \( m_1 = 12x^2 - 3 \), then the slope of the normal \( m_2 \) is:

\[ m_2 = -\frac{1}{m_1} = -\frac{1}{12x^2 - 3} \]

### Step 3: Find the slope of the given line
Next, we need to find the slope of the line given by the equation \( 9x - y + 5 = 0 \). We can rewrite this in slope-intercept form \( y = mx + b \):

\[ y = 9x + 5 \]

Thus, the slope \( m \) of the line is \( 9 \).

### Step 4: Set the slopes equal for perpendicularity
Since the normal line is perpendicular to the given line, we set the slope of the normal equal to the negative reciprocal of the slope of the line:

\[ -\frac{1}{12x^2 - 3} = -\frac{1}{9} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:

\[ 9 = 12x^2 - 3 \]

Rearranging this equation:

\[ 12x^2 = 12 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 6: Find corresponding \( y \) values
Now we substitute \( x = 1 \) and \( x = -1 \) back into the original curve equation to find the corresponding \( y \) values.

For \( x = 1 \):

\[ y = 4(1)^3 - 3(1) + 5 = 4 - 3 + 5 = 6 \]

For \( x = -1 \):

\[ y = 4(-1)^3 - 3(-1) + 5 = -4 + 3 + 5 = 4 \]

Thus, the points are \( (1, 6) \) and \( (-1, 4) \).

### Step 7: Write the equations of the normals
Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \):

1. For the point \( (1, 6) \):

\[ y - 6 = -\frac{1}{9}(x - 1) \]

Multiplying through by \( 9 \):

\[ 9y - 54 = -x + 1 \implies x + 9y - 55 = 0 \]

2. For the point \( (-1, 4) \):

\[ y - 4 = -\frac{1}{9}(x + 1) \]

Multiplying through by \( 9 \):

\[ 9y - 36 = -x - 1 \implies x + 9y - 35 = 0 \]

### Final Equations of the Normals
The equations of the normals are:

1. \( x + 9y - 55 = 0 \)
2. \( x + 9y - 35 = 0 \)