The point on the curve `y^2 = x`, the tangent at which makes an angle of 45° with x-axis will be given by

Options

A. `(1/2,1/4)`
B. `(1/2,1/2)`
C. `(2,4)`
D. `(1/4,1/2)`

Answer

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

### Step 1: Differentiate the curve
Given the curve equation:
\[ y^2 = x \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(x) \]
Using the chain rule, we have:
\[ 2y \frac{dy}{dx} = 1 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
From the differentiation, we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{2y} \]

### Step 3: Set the slope equal to the tangent of the angle
Since the tangent of the angle \( \theta \) that the tangent line makes with the x-axis is given by:
\[ \tan(\theta) = \frac{dy}{dx} \]
For \( \theta = 45^\circ \):
\[ \tan(45^\circ) = 1 \]
Thus, we set:
\[ \frac{1}{2y} = 1 \]

### Step 4: Solve for \( y \)
Now, we solve the equation:
\[ \frac{1}{2y} = 1 \implies 2y = 1 \implies y = \frac{1}{2} \]

### Step 5: Find the corresponding \( x \) value
Now that we have \( y \), we can find \( x \) using the original curve equation:
\[ y^2 = x \implies \left(\frac{1}{2}\right)^2 = x \implies x = \frac{1}{4} \]

### Step 6: State the point
Thus, the point on the curve where the tangent makes an angle of 45° with the x-axis is:
\[ \left(\frac{1}{4}, \frac{1}{2}\right) \]

### Final Answer
The point is \( \left(\frac{1}{4}, \frac{1}{2}\right) \).

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