The equation of a tangent to the parabola `y^2=""8x""i s""y""=""x""+""2` . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is (1) `(-1,""1)` (2) `(0,""2)` (3) `(2,""4)` (4) `(-2,""0)`

Options

A. (2, 4)
B. (-2, 0)
C. (-1, 1)
D. (0, 2)

Explanation

To solve the problem, we need to find the point on the line \( y = x + 2 \) from which the other tangent to the parabola \( y^2 = 8x \) is perpendicular to the given tangent.

### Step-by-step Solution:

1. **Identify the Parabola and the Tangent Line**:
- The equation of the parabola is \( y^2 = 8x \).
- The equation of the tangent line is \( y = x + 2 \).

2. **Find the Slope of the Given Tangent**:
- The slope of the line \( y = x + 2 \) is \( m = 1 \).

3. **Determine the Slope of the Perpendicular Tangent**:
- The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope.
- Therefore, the slope of the perpendicular tangent is \( m_{\perp} = -1 \).

4. **Equation of the Perpendicular Tangent**:
- The equation of the tangent to the parabola at a point \( (x_1, y_1) \) can be expressed as \( yy_1 = 4(x + x_1) \).
- Since we want this tangent to have a slope of \( -1 \), we can express the tangent line in point-slope form:
\[ y - y_1 = -1(x - x_1) \]

5. **Finding the Directrix of the Parabola**:
- The directrix of the parabola \( y^2 = 8x \) is given by the equation \( x = -2 \).

6. **Finding the Intersection of the Directrix and the Given Line**:
- To find the point of intersection of the line \( y = x + 2 \) and the directrix \( x = -2 \), substitute \( x = -2 \) into the line equation:
\[ y = -2 + 2 = 0 \]
- Thus, the point of intersection is \( (-2, 0) \).

### Conclusion:
The point on the line \( y = x + 2 \) from which the other tangent to the parabola \( y^2 = 8x \) is perpendicular to the given tangent is \( (-2, 0) \).