Find the point on the curve` y^(2) = x` at which the tangent drawn makes an angle of `45^(@)` from X-axis.

Answer

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

### Step 1: Understand the slope of the tangent
The slope of the tangent line at any point on the curve is given by the tangent of the angle it makes with the x-axis. For an angle of \( 45^\circ \), the slope \( m \) is:
\[ m = \tan(45^\circ) = 1 \]

### Step 2: Differentiate the curve
The given curve is \( y^2 = x \). To find the slope of the tangent at any point, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(x) \]
Using implicit differentiation, we have:
\[ 2y \frac{dy}{dx} = 1 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives us:
\[ \frac{dy}{dx} = \frac{1}{2y} \]

### Step 4: Set the slope equal to 1
Since we want the slope of the tangent to be 1, we set:
\[ \frac{1}{2y} = 1 \]

### Step 5: Solve for \(y\)
Cross-multiplying gives:
\[ 1 = 2y \implies y = \frac{1}{2} \]

### Step 6: Find the corresponding \(x\)
Now that we have \(y\), we can find \(x\) using the original curve equation \(y^2 = x\):
\[ x = y^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 7: State the point
Thus, the point on the curve where the tangent makes an angle of \( 45^\circ \) with the x-axis is:
\[ \left( \frac{1}{4}, \frac{1}{2} \right) \]

### Summary
The point on the curve \( y^2 = x \) at which the tangent makes an angle of \( 45^\circ \) with the x-axis is \( \left( \frac{1}{4}, \frac{1}{2} \right) \).

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