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: Understand the Parabola
The given parabola is \( y^2 = 2x \). This is a standard form of a parabola that opens to the right. The vertex of this parabola is at the origin (0,0).

### Step 2: Determine the Slope of the Tangent
A tangent inclined at \( 45^\circ \) to the x-axis has a slope \( m = \tan(45^\circ) = 1 \) or \( m = \tan(135^\circ) = -1 \). Thus, we have two possible slopes for the tangent line.

### Step 3: Use the Point-Slope Form of the Tangent
The equation of the tangent to the parabola \( y^2 = 2x \) at a point \( (x_1, y_1) \) can be expressed as:
\[ yy_1 = x + x_1 \]
where \( y_1^2 = 2x_1 \).

### Step 4: Substitute the Slope into the Tangent Equation
For the slope \( m = 1 \):
\[ y - y_1 = 1(x - x_1) \]
This can be rearranged to:
\[ y = x - x_1 + y_1 \]

For the slope \( m = -1 \):
\[ y - y_1 = -1(x - x_1) \]
This can be rearranged to:
\[ y = -x + x_1 + y_1 \]

### Step 5: Solve for the Point of Contact
We will solve for the point of contact using the slope \( m = 1 \) first. Substituting \( y = x - x_1 + y_1 \) into the parabola equation:
\[ (x - x_1 + y_1)^2 = 2x \]
Expanding and rearranging gives us a quadratic equation in \( x \).

### Step 6: Find the Values of \( x_1 \) and \( y_1 \)
Using the condition \( y_1^2 = 2x_1 \) and substituting back into the quadratic equation, we can find the values of \( x_1 \) and \( y_1 \).

### Step 7: Repeat for the Second Slope
Repeat the process for the slope \( m = -1 \) to find the corresponding point of contact.

### Step 8: Conclusion
After solving both cases, we will find the points of contact for both slopes. The correct point of contact will be the one that satisfies the conditions of the problem.

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