The point on the curve `y^(2)=x` where tangent makes `45^(@)` angle with x-axis, is

Options

A. `x-y=-(1)/(4)`
B. `x-y=(1)/(4)`
C. `x+y=(1)/(2)`
D. `x-y=(1)/(2)`

Answer

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

### Step 1: Understand the slope of the tangent
The slope of the tangent line that makes a \( 45^\circ \) angle with the x-axis is given by:
\[ m = \tan(45^\circ) = 1 \]

### Step 2: Differentiate the curve
Given the curve \( y^2 = x \), we need to find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent at any point on the curve. We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(x) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 1 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2y} \]

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

### Step 5: Solve for \( y \)
Multiplying both sides by \( 2y \) gives:
\[ 1 = 2y \implies y = \frac{1}{2} \]

### Step 6: Find the corresponding \( x \)
Now, we substitute \( y = \frac{1}{2} \) back into the original curve equation \( y^2 = x \) to find \( x \):
\[ \left(\frac{1}{2}\right)^2 = x \implies x = \frac{1}{4} \]

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

### Final Answer
The point on the curve \( y^2 = x \) where the tangent makes a \( 45^\circ \) angle with the x-axis is:
\[ \left(\frac{1}{4}, \frac{1}{2}\right) \]