The line `x+y=6` is normal to the parabola `y^(2)=8x` at the point.

Options

A. (4,2)
B. (2,4)
C. (2,2)
D. (3,3)

Explanation

To find the point on the parabola \( y^2 = 8x \) where the line \( x + y = 6 \) is normal, we can follow these steps:

### Step 1: Rewrite the line equation
The line equation \( x + y = 6 \) can be rewritten in slope-intercept form:
\[ y = -x + 6 \]
**Hint:** Rearranging the equation helps to identify the slope of the line.

### Step 2: Identify the parabola
The given parabola is \( y^2 = 8x \). We can identify that it is in the standard form \( y^2 = 4ax \) where \( 4a = 8 \), thus \( a = 2 \).
**Hint:** Recognizing the standard form of the parabola helps in deriving properties related to its normals.

### Step 3: Use the normal line equation
The equation of the normal to the parabola \( y^2 = 4ax \) at the point \( (at^2, 2at) \) is given by:
\[ y = mx - 2am - am^3 \]
where \( m \) is the slope of the normal line. Since the line \( y = -x + 6 \) has a slope of \( -1 \), we set \( m = -1 \).

**Hint:** The slope of the normal line is crucial for determining the point of tangency on the parabola.

### Step 4: Substitute \( m \) into the normal equation
Substituting \( m = -1 \) and \( a = 2 \) into the normal equation:
\[ y = -x - 2(2)(-1) - 2(-1)^3 \]
This simplifies to:
\[ y = -x + 4 + 2 = -x + 6 \]
**Hint:** Confirming the equation matches the given line ensures that we are on the right track.

### Step 5: Find the point of tangency
The point on the parabola can be derived from the normal line equation. The coordinates of the point on the parabola where the normal intersects can be expressed as:
\[ (at^2, 2at) = (2t^2, 4t) \]
We know the normal line intersects at \( (2t^2, 4t) \) and should satisfy the line equation \( x + y = 6 \):
\[ 2t^2 + 4t = 6 \]
This simplifies to:
\[ 2t^2 + 4t - 6 = 0 \]
**Hint:** Setting up the equation based on the intersection condition is essential for finding the correct point.

### Step 6: Solve the quadratic equation
Dividing the entire equation by 2:
\[ t^2 + 2t - 3 = 0 \]
Factoring gives:
\[ (t + 3)(t - 1) = 0 \]
Thus, \( t = -3 \) or \( t = 1 \).

**Hint:** Solving the quadratic equation will yield potential values for \( t \).

### Step 7: Find corresponding points
Using \( t = 1 \):
\[ x = 2(1^2) = 2, \quad y = 4(1) = 4 \]
Using \( t = -3 \):
\[ x = 2(-3)^2 = 18, \quad y = 4(-3) = -12 \]
The points are \( (2, 4) \) and \( (18, -12) \).

**Hint:** Evaluating both values of \( t \) gives all possible points where the normal can touch the parabola.

### Conclusion
The point on the parabola where the line \( x + y = 6 \) is normal is:
\[ \boxed{(2, 4)} \]