Find the equation of the normal to the parabola `y^(2)=4x`, which is
(a) parallel to the line y = 2x - 5
(b) perpendicular to the line x + 3y + 1 = 0.

Answer

To find the equation of the normal to the parabola \( y^2 = 4x \) that meets the specified conditions, we will follow these steps:

### Step 1: Identify the parabola and its parameters
The given parabola is \( y^2 = 4x \). Here, \( a = 1 \) (since \( 4a = 4 \)).

### Step 2: Write the equation of the normal line
The equation of the normal to the parabola at a point \( (at^2, 2at) \) is given by:
\[ y = mx - 2am - am^3 \]
where \( m \) is the slope of the normal.

### Part (a): Normal parallel to the line \( y = 2x - 5 \)

#### Step 3: Find the slope of the given line
The slope of the line \( y = 2x - 5 \) is \( 2 \). Therefore, we set \( m = 2 \).

#### Step 4: Substitute \( m \) into the normal equation
Substituting \( m = 2 \) into the normal equation:
\[ y = 2x - 2(1)(2) - (1)(2^3) \]
\[ y = 2x - 4 - 8 \]
\[ y = 2x - 12 \]

### Part (b): Normal perpendicular to the line \( x + 3y + 1 = 0 \)

#### Step 5: Find the slope of the given line
Rearranging the line \( x + 3y + 1 = 0 \) into slope-intercept form:
\[ 3y = -x - 1 \implies y = -\frac{1}{3}x - \frac{1}{3} \]
The slope of this line is \( -\frac{1}{3} \).

#### Step 6: Find the slope of the normal
For two lines to be perpendicular, the product of their slopes must be \( -1 \). Let the slope of the normal be \( m_2 \):
\[ m_1 \cdot m_2 = -1 \implies -\frac{1}{3} \cdot m_2 = -1 \implies m_2 = 3 \]

#### Step 7: Substitute \( m_2 \) into the normal equation
Substituting \( m = 3 \) into the normal equation:
\[ y = 3x - 2(1)(3) - (1)(3^3) \]
\[ y = 3x - 6 - 27 \]
\[ y = 3x - 33 \]

### Final Answers:
- (a) The equation of the normal parallel to the line \( y = 2x - 5 \) is \( y = 2x - 12 \).
- (b) The equation of the normal perpendicular to the line \( x + 3y + 1 = 0 \) is \( y = 3x - 33 \).