Find the equation of the normal to the curve : `y=x^(3)+5x^(2)-10x+10,`
where the normal is parallel to the line `x-2y+10=0.`

Answer

To find the equation of the normal to the curve \( y = x^3 + 5x^2 - 10x + 10 \) where the normal is parallel to the line \( x - 2y + 10 = 0 \), we will follow these steps:

### Step 1: Find the slope of the given line
The equation of the line can be rearranged to slope-intercept form \( y = mx + b \):

\[ x - 2y + 10 = 0 \implies 2y = x + 10 \implies y = \frac{1}{2}x + 5 \]

Thus, the slope \( m \) of the line is \( \frac{1}{2} \).

### Step 2: Find the derivative of the curve
Next, we differentiate the curve to find the slope of the tangent line:

\[ y = x^3 + 5x^2 - 10x + 10 \]

Differentiating with respect to \( x \):

\[ \frac{dy}{dx} = 3x^2 + 10x - 10 \]

### Step 3: Set up the equation for the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. If the slope of the tangent is \( m_t \), then:

\[ m_n = -\frac{1}{m_t} = -\frac{1}{3x^2 + 10x - 10} \]

We want the normal to be parallel to the given line, which has a slope of \( \frac{1}{2} \). Therefore, we set:

\[ -\frac{1}{3x^2 + 10x - 10} = \frac{1}{2} \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:

\[ -2 = 3x^2 + 10x - 10 \]

Rearranging this equation:

\[ 3x^2 + 10x + 8 = 0 \]

### Step 5: Factor or use the quadratic formula
We can factor this quadratic equation:

\[ (3x + 4)(x + 2) = 0 \]

Thus, the solutions for \( x \) are:

\[ x = -\frac{4}{3} \quad \text{and} \quad x = -2 \]

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

1. For \( x = -\frac{4}{3} \):

\[ y = \left(-\frac{4}{3}\right)^3 + 5\left(-\frac{4}{3}\right)^2 - 10\left(-\frac{4}{3}\right) + 10 \]
Calculating each term:
\[ = -\frac{64}{27} + 5 \cdot \frac{16}{9} + \frac{40}{3} + 10 \]
\[ = -\frac{64}{27} + \frac{80}{9} + \frac{360}{27} + \frac{270}{27} \]
\[ = -\frac{64}{27} + \frac{80 \cdot 3}{27} + \frac{360}{27} + \frac{270}{27} \]
\[ = -\frac{64}{27} + \frac{240}{27} + \frac{360}{27} + \frac{270}{27} \]
\[ = \frac{240 + 360 + 270 - 64}{27} = \frac{806}{27} \]

So, the point is \( \left(-\frac{4}{3}, \frac{806}{27}\right) \).

2. For \( x = -2 \):

\[ y = (-2)^3 + 5(-2)^2 - 10(-2) + 10 \]
Calculating:
\[ = -8 + 20 + 20 + 10 = 42 \]

So, the point is \( (-2, 42) \).

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

1. For the point \( \left(-\frac{4}{3}, \frac{806}{27}\right) \):

\[ y - \frac{806}{27} = \frac{1}{2}\left(x + \frac{4}{3}\right) \]

2. For the point \( (-2, 42) \):

\[ y - 42 = \frac{1}{2}(x + 2) \]

### Final Equations of the Normals
1. From \( \left(-\frac{4}{3}, \frac{806}{27}\right) \):
\[ y = \frac{1}{2}x + \frac{806}{27} - \frac{2}{3} \]

2. From \( (-2, 42) \):
\[ y = \frac{1}{2}x + 43 \]