Find the point on the curve `y^2=a x` the tangent at which makes an angle of 45^0 with the x-axis.

Explanation

To find the point on the curve \( y^2 = ax \) where the tangent makes an angle of \( 45^\circ \) with the x-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Curve**: The given curve is \( y^2 = ax \). This is a standard form of a parabola that opens to the right.

2. **Find the Slope of the Tangent**: The slope \( m \) of the tangent line that makes an angle of \( 45^\circ \) with the x-axis is given by:
\[ m = \tan(45^\circ) = 1 \]

3. **Equation of the Tangent Line**: The equation of the tangent to the curve \( y^2 = ax \) at the point \( (x_0, y_0) \) can be expressed as:
\[ yy_0 = \frac{a}{2}(x + x_0) \]
where \( (x_0, y_0) \) is the point of tangency.

4. **Substituting the Slope**: Since we know that the slope \( m = 1 \), we can express the tangent line in point-slope form:
\[ y - y_0 = m(x - x_0) \implies y - y_0 = (x - x_0) \]
Rearranging gives:
\[ y = x - x_0 + y_0 \]

5. **Finding the Point of Tangency**: The point \( (x_0, y_0) \) lies on the curve, so we have:
\[ y_0^2 = ax_0 \]

6. **Using the Slope in the Tangent Equation**: From the tangent equation \( yy_0 = \frac{a}{2}(x + x_0) \) and substituting \( m = 1 \):
\[ y = x - x_0 + y_0 \implies y_0(y - y_0) = \frac{a}{2}(x + x_0) \]

7. **Setting up the Equation**: Since \( y_0 = \sqrt{ax_0} \), we can substitute \( y_0 \) into the equation:
\[ \sqrt{ax_0}(y - \sqrt{ax_0}) = \frac{a}{2}(x + x_0) \]

8. **Solving for \( x_0 \) and \( y_0 \)**: We can simplify this equation to find \( x_0 \):
\[ \sqrt{ax_0} \cdot (x - x_0) = \frac{a}{2}(x + x_0) \]
After some algebra, we can solve for \( x_0 \) and \( y_0 \).

9. **Final Calculation**: After substituting and simplifying, we find:
\[ x_0 = \frac{a}{4}, \quad y_0 = \frac{a}{2} \]

10. **Result**: The point on the curve where the tangent makes an angle of \( 45^\circ \) with the x-axis is:
\[ \left( \frac{a}{4}, \frac{a}{2} \right) \]