The point of contact of the tangent of `y^2=2x` inclined to `45^(@)` to the axis of the parabola is :

Options

A. `((-1)/(2),0)`
B. `(1,pm2)`
C. `((1)/(2),pm1)`
D. None of these

Answer

To find the point of contact of the tangent to the parabola \( y^2 = 2x \) that is inclined at \( 45^\circ \) to the axis of the parabola, we can follow these steps:

### Step 1: Identify the standard form of the parabola
The given parabola is \( y^2 = 2x \). This can be compared to the standard form \( y^2 = 4ax \), where \( a = \frac{1}{2} \).

### Step 2: Determine the slope of the tangent
The slope \( m \) of a line inclined at \( 45^\circ \) to the x-axis is given by:
\[ m = \tan(45^\circ) = 1 \]
Thus, the slope of the tangent line to the parabola is \( m = 1 \).

### Step 3: Use the point-slope form of the tangent line
The equation of the tangent line to the parabola at the point \( (x_0, y_0) \) can be expressed as:
\[ y - y_0 = m(x - x_0) \]
Substituting \( m = 1 \), we have:
\[ y - y_0 = (x - x_0) \]

### Step 4: Find the point of contact
For the parabola \( y^2 = 2x \), the coordinates of the point of contact can be expressed in terms of a parameter \( t \):
\[ x_0 = \frac{1}{2}t^2, \quad y_0 = t \]
Substituting these into the tangent equation gives:
\[ t - t_0 = (x - \frac{1}{2}t^2) \]

### Step 5: Find the derivative
To find the slope of the tangent line at the point \( (x_0, y_0) \), we differentiate \( y^2 = 2x \):
\[ 2y \frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y} \]
At the point of contact, this slope must equal \( 1 \):
\[ \frac{1}{y_0} = 1 \implies y_0 = 1 \text{ or } y_0 = -1 \]

### Step 6: Solve for \( t \)
Since \( y_0 = t \), we have:
\[ t = 1 \text{ or } t = -1 \]

### Step 7: Calculate the corresponding \( x_0 \)
Using \( t = 1 \):
\[ x_0 = \frac{1}{2}(1)^2 = \frac{1}{2} \]
Using \( t = -1 \):
\[ x_0 = \frac{1}{2}(-1)^2 = \frac{1}{2} \]

### Step 8: Write the points of contact
Thus, the points of contact are:
\[ \left(\frac{1}{2}, 1\right) \quad \text{and} \quad \left(\frac{1}{2}, -1\right) \]

### Final Answer
The points of contact of the tangent to the parabola \( y^2 = 2x \) inclined at \( 45^\circ \) to the axis are:
\[ \left(\frac{1}{2}, 1\right) \text{ and } \left(\frac{1}{2}, -1\right) \]