The equation of common tangent to the curve `y^2=4x` and `xy=-1` is

Options

A. `2^(2/3)*y=x+2^(4/3)`
B. `2^(2/3)*x=y+2^(4/3)`
C. `y=2^(2/3)*x-2^(4/3)`
D. `y=x-2^(2/3)`

Answer

To find the equation of the common tangent to the curves \(y^2 = 4x\) and \(xy = -1\), we can follow these steps:

### Step 1: Identify the Tangent Equation for the Parabola
The equation of the parabola \(y^2 = 4x\) can be expressed in the standard form \(y^2 = 4ax\) where \(a = 1\). The equation of the tangent to this parabola at a point can be given by:
\[ y = mx + \frac{1}{m} \]
where \(m\) is the slope of the tangent.

### Step 2: Substitute the Tangent Equation into the Second Curve
The second curve is given by \(xy = -1\). We can substitute \(y\) from the tangent equation into this equation:
\[ x\left(mx + \frac{1}{m}\right) = -1 \]
This simplifies to:
\[ mx^2 + \frac{x}{m} + 1 = 0 \]

### Step 3: Set the Discriminant to Zero
For the tangent to be common, the quadratic equation must have exactly one solution, which means the discriminant must be zero. The discriminant \(D\) of the quadratic \(Ax^2 + Bx + C = 0\) is given by:
\[ D = B^2 - 4AC \]
Here, \(A = m\), \(B = \frac{1}{m}\), and \(C = 1\). Thus, the discriminant becomes:
\[ D = \left(\frac{1}{m}\right)^2 - 4m \cdot 1 = \frac{1}{m^2} - 4m \]
Setting the discriminant to zero gives:
\[ \frac{1}{m^2} - 4m = 0 \]

### Step 4: Solve for \(m\)
Multiplying through by \(m^2\) (assuming \(m \neq 0\)):
\[ 1 - 4m^3 = 0 \]
This implies:
\[ 4m^3 = 1 \quad \Rightarrow \quad m^3 = \frac{1}{4} \quad \Rightarrow \quad m = \frac{1}{\sqrt[3]{4}} = \frac{1}{2\sqrt[3]{2}} \]

### Step 5: Write the Equation of the Tangent
Now, substituting \(m\) back into the tangent equation:
\[ y = mx + \frac{1}{m} \]
Calculating \(\frac{1}{m}\):
\[ \frac{1}{m} = 2\sqrt[3]{2} \]
Thus, the equation of the tangent becomes:
\[ y = \frac{1}{2\sqrt[3]{2}}x + 2\sqrt[3]{2} \]

### Step 6: Rearranging the Equation
To express it in a standard form, we can rearrange:
\[ y - \frac{1}{2\sqrt[3]{2}}x - 2\sqrt[3]{2} = 0 \]
This represents the equation of the common tangent.

### Final Answer
The equation of the common tangent to the curves \(y^2 = 4x\) and \(xy = -1\) is:
\[ y = \frac{1}{2\sqrt[3]{2}}x + 2\sqrt[3]{2} \]