Find the equation of normal to parabola `y=x^(2)-3x-4`
(a) at point (3,-4)
(b) having slope 5.

Explanation

To find the equations of the normal to the parabola \( y = x^2 - 3x - 4 \) at the specified conditions, we will solve the problem in two parts.

### Part (a): Find the equation of the normal at the point (3, -4)

1. **Identify the point on the parabola**: The given point is \( P(3, -4) \).

2. **Differentiate the parabola**: We need to find the derivative \( \frac{dy}{dx} \) to determine the slope of the tangent line at point \( P \).
\[ \frac{dy}{dx} = 2x - 3 \]

3. **Evaluate the derivative at \( x = 3 \)**:
\[ \frac{dy}{dx} \bigg|_{x=3} = 2(3) - 3 = 6 - 3 = 3 \]
Thus, the slope of the tangent at point \( P \) is \( 3 \).

4. **Find the slope of the normal**: The slope of the normal is the negative reciprocal of the slope of the tangent.
\[ \text{slope of normal} = -\frac{1}{3} \]

5. **Use the point-slope form to find the equation of the normal**:
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) \) is the point \( P(3, -4) \) and \( m \) is the slope of the normal.
\[ y - (-4) = -\frac{1}{3}(x - 3) \]
Simplifying this:
\[ y + 4 = -\frac{1}{3}x + 1 \]
\[ y = -\frac{1}{3}x + 1 - 4 \]
\[ y = -\frac{1}{3}x - 3 \]

6. **Final equation of the normal**:
\[ y = -\frac{1}{3}x - 3 \]

### Part (b): Find the equation of the normal with slope 5

1. **Identify the slope of the normal**: Given that the slope of the normal is \( 5 \), we know that:
\[ -\frac{dx}{dy} = 5 \implies \frac{dx}{dy} = -5 \implies \frac{dy}{dx} = -\frac{1}{5} \]

2. **Set up the equation for the slope of the tangent**:
\[ 2x - 3 = -\frac{1}{5} \]

3. **Solve for \( x \)**:
\[ 2x = 3 - \frac{1}{5} \]
To combine the terms, convert \( 3 \) to a fraction:
\[ 3 = \frac{15}{5} \implies 2x = \frac{15}{5} - \frac{1}{5} = \frac{14}{5} \]
\[ x = \frac{14}{10} = \frac{7}{5} \]

4. **Find the corresponding \( y \) value**:
Substitute \( x = \frac{7}{5} \) back into the parabola equation:
\[ y = \left(\frac{7}{5}\right)^2 - 3\left(\frac{7}{5}\right) - 4 \]
\[ y = \frac{49}{25} - \frac{21}{5} - 4 \]
Convert \( \frac{21}{5} \) and \( 4 \) to have a common denominator:
\[ \frac{21}{5} = \frac{105}{25}, \quad 4 = \frac{100}{25} \]
\[ y = \frac{49}{25} - \frac{105}{25} - \frac{100}{25} = \frac{49 - 105 - 100}{25} = \frac{-156}{25} \]

5. **Use the point-slope form to find the equation of the normal**:
The point is \( P\left(\frac{7}{5}, -\frac{156}{25}\right) \) and the slope is \( 5 \):
\[ y - \left(-\frac{156}{25}\right) = 5\left(x - \frac{7}{5}\right) \]
Simplifying:
\[ y + \frac{156}{25} = 5x - 7 \]
\[ y = 5x - 7 - \frac{156}{25} \]
Convert \( -7 \) to a fraction:
\[ -7 = -\frac{175}{25} \]
\[ y = 5x - \frac{175 + 156}{25} = 5x - \frac{31}{25} \]

6. **Final equation of the normal**:
\[ y = 5x - \frac{31}{25} \]