From a variable point P on the tagent at the vertex of the parabola `y^(2)=2x`, a line is drawn perpendicular to the chord of contact. These variable lines always pass through a fixed point, whose x - coordinate is

Options

A. `(1)/(2)`
B. 1
C. `(3)/(2)`
D. 2

Answer

To solve the problem step by step, we will analyze the given parabola and the conditions provided in the question.

### Step 1: Identify the Parabola
The equation of the parabola is given as:
\[ y^2 = 2x \]
This parabola opens to the right with its vertex at the origin (0, 0).

**Hint:** Remember that the vertex of the parabola \( y^2 = 2x \) is at (0, 0).

### Step 2: Determine the Tangent at the Vertex
The tangent to the parabola at the vertex (0, 0) is the x-axis, which can be represented as:
\[ y = 0 \]

**Hint:** The tangent at the vertex of a parabola is a horizontal line when the vertex is at the origin.

### Step 3: Define the Variable Point P
Let the variable point \( P \) on the tangent line be defined as:
\[ P(0, k) \]
where \( k \) is any real number.

**Hint:** Since point \( P \) lies on the x-axis (tangent at the vertex), its x-coordinate is always 0.

### Step 4: Find the Chord of Contact
The chord of contact from point \( P(0, k) \) can be derived using the formula for the chord of contact for the parabola \( y^2 = 4ax \). Here, \( a = \frac{1}{2} \):
\[ yy_1 = 2(x + x_1) \]
Substituting \( x_1 = 0 \) and \( y_1 = k \):
\[ yk = 2(x + 0) \]
This simplifies to:
\[ yk = 2x \]
or
\[ x = \frac{ky}{2} \]

**Hint:** The chord of contact is derived from the point from which the tangents are drawn to the parabola.

### Step 5: Determine the Slope of the Chord of Contact
From the equation \( x = \frac{ky}{2} \), we can express it in slope-intercept form:
\[ y = \frac{2}{k}x \]
Thus, the slope \( m_{COC} \) of the chord of contact is:
\[ m_{COC} = \frac{2}{k} \]

**Hint:** The slope of a line in the form \( y = mx + b \) is simply \( m \).

### Step 6: Find the Slope of the Perpendicular Line
The slope of the line perpendicular to the chord of contact is given by:
\[ m_{perpendicular} = -\frac{1}{m_{COC}} = -\frac{1}{\frac{2}{k}} = -\frac{k}{2} \]

**Hint:** The product of the slopes of two perpendicular lines is -1.

### Step 7: Write the Equation of the Perpendicular Line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( P(0, k) \) and the slope \( -\frac{k}{2} \):
\[ y - k = -\frac{k}{2}(x - 0) \]
This simplifies to:
\[ y - k = -\frac{k}{2}x \]
or
\[ y = -\frac{k}{2}x + k \]

**Hint:** The point-slope form is useful for finding the equation of a line given a point and a slope.

### Step 8: Analyze the Family of Lines
The equation \( y = -\frac{k}{2}x + k \) represents a family of lines parameterized by \( k \). We need to find the fixed point through which all these lines pass.

**Hint:** A family of lines can be analyzed by eliminating the parameter \( k \).

### Step 9: Set Up the Condition for Fixed Point
To find the fixed point, we can set \( y = 0 \) in the line equation:
\[ 0 = -\frac{k}{2}x + k \]
Rearranging gives:
\[ \frac{k}{2}x = k \]
Assuming \( k \neq 0 \):
\[ x = 2 \]

**Hint:** Setting \( y = 0 \) helps find the x-coordinate of the fixed point.

### Step 10: Conclusion
Thus, the x-coordinate of the fixed point through which all the variable lines pass is:
\[ \text{x-coordinate} = 2 \]

**Final Answer:** The x-coordinate of the fixed point is \( 2 \).