The common tangent of the curves `y=x^(2) +(1)/(x) " and " y^(2) =4 x` is

Options

A. `y=x +1`
B. `y=x-1`
C. `y=-x+1`
D. `y=-x-1`

Answer

To find the common tangent of the curves \( y = x^2 + \frac{1}{x} \) and \( y^2 = 4x \), we can follow these steps:

### Step 1: Identify the curves
The two curves given are:
1. \( y = x^2 + \frac{1}{x} \) (Curve 1)
2. \( y^2 = 4x \) (Curve 2)

### Step 2: Find a point of intersection
To find a common tangent, we first need to find a point where both curves intersect. We can do this by substituting \( y \) from Curve 1 into Curve 2.

From Curve 2, we have:
\[ y^2 = 4x \]
Substituting \( y = x^2 + \frac{1}{x} \) into this equation:
\[ \left(x^2 + \frac{1}{x}\right)^2 = 4x \]

Expanding the left-hand side:
\[ x^4 + 2 + \frac{1}{x^2} = 4x \]

Rearranging gives:
\[ x^4 - 4x + 2 + \frac{1}{x^2} = 0 \]

To eliminate the fraction, multiply through by \( x^2 \):
\[ x^6 - 4x^3 + 2x^2 + 1 = 0 \]

### Step 3: Check for specific values
We can check specific values of \( x \) to find intersections. Let's try \( x = 1 \):
\[ y = 1^2 + \frac{1}{1} = 2 \]
For Curve 2:
\[ y^2 = 4 \cdot 1 \implies y = 2 \]
Thus, the point of intersection is \( (1, 2) \).

### Step 4: Find the slopes of the curves at the intersection point
Now we need to find the derivatives of both curves at the point \( (1, 2) \).

**For Curve 1:**
\[ y = x^2 + \frac{1}{x} \]
Differentiating:
\[ \frac{dy}{dx} = 2x - \frac{1}{x^2} \]
At \( x = 1 \):
\[ \frac{dy}{dx} = 2(1) - 1 = 1 \]

**For Curve 2:**
\[ y^2 = 4x \]
Differentiating implicitly:
\[ 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]
At \( (1, 2) \):
\[ \frac{dy}{dx} = \frac{2}{2} = 1 \]

### Step 5: Write the equation of the common tangent
Since both derivatives at the point \( (1, 2) \) are equal (both are 1), the slope of the common tangent is 1. The equation of the tangent line can be written using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (1, 2) \) and \( m = 1 \):
\[ y - 2 = 1(x - 1) \]
Simplifying gives:
\[ y - 2 = x - 1 \implies y = x + 1 \]

### Final Answer
The equation of the common tangent is:
\[ \boxed{y = x + 1} \]