Line `x-y=1` intersect the parabola `y^(2)=4x` at A and B. Normals at A and B intersect at C. If D is the point at which line CD is normal to the parabola, then coordinate of point D is

Options

A. `(4,-4)`
B. (4,4)
C. `(-4,-4)`
D. None of these

Explanation

To solve the problem, we need to find the coordinates of point D where the line CD is normal to the parabola \( y^2 = 4x \). Here’s a step-by-step solution:

### Step 1: Find Points of Intersection A and B
We start with the line \( x - y = 1 \) and the parabola \( y^2 = 4x \).

1. Rearranging the line equation gives us:
\[ x = y + 1 \]

2. Substitute \( x \) in the parabola's equation:
\[ y^2 = 4(y + 1) \]
\[ y^2 = 4y + 4 \]
\[ y^2 - 4y - 4 = 0 \]

### Step 2: Solve the Quadratic Equation
Now we solve the quadratic equation \( y^2 - 4y - 4 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = -4 \):
\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ y = \frac{4 \pm \sqrt{16 + 16}}{2} \]
\[ y = \frac{4 \pm \sqrt{32}}{2} \]
\[ y = \frac{4 \pm 4\sqrt{2}}{2} \]
\[ y = 2 \pm 2\sqrt{2} \]

### Step 3: Find Corresponding x-coordinates
Now, we find the corresponding x-coordinates for points A and B:
1. For \( y_1 = 2 + 2\sqrt{2} \):
\[ x_1 = (2 + 2\sqrt{2}) + 1 = 3 + 2\sqrt{2} \]
So, point A is \( A(3 + 2\sqrt{2}, 2 + 2\sqrt{2}) \).

2. For \( y_2 = 2 - 2\sqrt{2} \):
\[ x_2 = (2 - 2\sqrt{2}) + 1 = 3 - 2\sqrt{2} \]
So, point B is \( B(3 - 2\sqrt{2}, 2 - 2\sqrt{2}) \).

### Step 4: Find Normals at A and B
The slope of the tangent to the parabola \( y^2 = 4x \) at a point \( (x_0, y_0) \) is given by:
\[ \text{slope} = \frac{2y_0}{4} = \frac{y_0}{2} \]
Thus, the slope of the normal is:
\[ \text{slope of normal} = -\frac{2}{y_0} \]

1. For point A:
\[ y_0 = 2 + 2\sqrt{2} \Rightarrow \text{slope of normal at A} = -\frac{2}{2 + 2\sqrt{2}} \]

2. For point B:
\[ y_0 = 2 - 2\sqrt{2} \Rightarrow \text{slope of normal at B} = -\frac{2}{2 - 2\sqrt{2}} \]

### Step 5: Find Intersection Point C
The equations of the normals can be derived using point-slope form. After finding the equations, we can solve them simultaneously to find the coordinates of point C.

### Step 6: Find Point D
Point D lies on the parabola and is such that line CD is normal to the parabola. We can find the coordinates of D by using the properties of normals to the parabola.

1. The y-coordinate of D can be determined using the property that the sum of the y-coordinates of the normals at A, B, and D equals zero:
\[ (2 + 2\sqrt{2}) + (2 - 2\sqrt{2}) + y_D = 0 \]
\[ 4 + y_D = 0 \Rightarrow y_D = -4 \]

2. Substitute \( y_D = -4 \) into the parabola equation:
\[ (-4)^2 = 4x \Rightarrow 16 = 4x \Rightarrow x = 4 \]

### Final Coordinates of Point D
Thus, the coordinates of point D are:
\[ D(4, -4) \]