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 solve the problem, we need to find the point at which the line \( x + y = 6 \) is normal to the parabola \( y^2 = 8x \).

### Step-by-Step Solution:

1. **Rewrite the Line Equation**:
The line equation \( x + y = 6 \) can be rewritten in slope-intercept form:
\[ y = -x + 6 \]
The slope \( m \) of this line is \( -1 \).

**Hint**: To find the slope of a line given in standard form \( Ax + By + C = 0 \), rearrange it to the form \( y = mx + c \).

2. **Identify the Parabola**:
The given parabola is \( y^2 = 8x \). This can be compared to the standard form of a parabola \( y^2 = 4ax \). Here, we find:
\[ 4a = 8 \implies a = 2 \]

**Hint**: To find the parameter \( a \) of a parabola, compare the coefficients with the standard form \( y^2 = 4ax \).

3. **Equation of the Normal to the Parabola**:
The equation of the normal to the parabola at a point \( (am^2, -2am) \) is used. Here, \( m \) is the slope of the normal, which is \( -1 \).

Calculate the coordinates:
\[ x = am^2 = 2(-1)^2 = 2 \quad \text{and} \quad y = -2am = -2(2)(-1) = 4 \]

Thus, the point of tangency is \( (2, 4) \).

**Hint**: The coordinates of the point on the parabola where the normal line intersects can be derived using the formula for the normal.

4. **Verification**:
To verify, substitute \( x = 2 \) into the parabola equation:
\[ y^2 = 8(2) = 16 \implies y = \pm 4 \]
Since we found \( y = 4 \), the point \( (2, 4) \) lies on the parabola.

**Hint**: Always check if the point found lies on the original curve by substituting back into the equation.

5. **Conclusion**:
The line \( x + y = 6 \) is normal to the parabola \( y^2 = 8x \) at the point \( (2, 4) \).

**Final Answer**: The point is \( (2, 4) \).