The normal to parabola `y^(2) =4ax` from the point `(5a, -2a)` are

Options

A. `y = -x + 3a `
B. `y = 2x - 12a`
C. `y = 3x + 33a`
D. `y =x + 3a`

Answer

To find the normals to the parabola \( y^2 = 4ax \) from the point \( (5a, -2a) \), we will follow these steps:

### Step 1: Write the equation of the normal to the parabola in parametric form.
The equation of the normal to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ y + 2at = -\frac{1}{t}(x - at^2) \]
This can be rearranged to:
\[ y + 2at = -\frac{1}{t}x + at \]
or
\[ y + \frac{1}{t}x = at - 2at \]
Thus, the equation becomes:
\[ y + \frac{1}{t}x = -at \]

### Step 2: Substitute the point \( (5a, -2a) \) into the normal equation.
Substituting \( x = 5a \) and \( y = -2a \) into the normal equation:
\[ -2a + \frac{1}{t}(5a) = -at \]
This simplifies to:
\[ -2a + \frac{5a}{t} = -at \]

### Step 3: Rearrange the equation.
Rearranging gives:
\[ \frac{5a}{t} + at = 2a \]
Multiplying through by \( t \) to eliminate the fraction:
\[ 5a + at^2 = 2at \]
Rearranging this yields:
\[ at^2 - 2at + 5a = 0 \]

### Step 4: Factor out \( a \) (assuming \( a \neq 0 \)).
Dividing through by \( a \):
\[ t^2 - 2t + 5 = 0 \]

### Step 5: Find the roots of the quadratic equation.
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = 5 \):
\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 - 20}}{2} \]
\[ t = \frac{2 \pm \sqrt{-16}}{2} \]
\[ t = \frac{2 \pm 4i}{2} \]
\[ t = 1 \pm 2i \]

### Step 6: Write the equations of the normals.
1. For \( t = 1 + 2i \):
\[ y + 2a(1 + 2i) = -\frac{1}{1 + 2i}(x - a(1 + 2i)^2) \]
Calculate \( (1 + 2i)^2 = 1 + 4i - 4 = -3 + 4i \):
\[ y + 2a + 4ai = -\frac{1}{1 + 2i}(x - a(-3 + 4i)) \]
This gives us the first normal.

2. For \( t = 1 - 2i \):
Similar calculations will yield the second normal.

### Conclusion:
The normals to the parabola \( y^2 = 4ax \) from the point \( (5a, -2a) \) are given by the equations derived above.