Equation of a common tangent to the parabola `y^(2)=4x` and the hyperbola xy=2 is

Options

A. x+2y+4=0
B. x-2y+4=0
C. x+y+1=0
D. 4x+2y+1=0

Answer

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

### Step 1: Identify the parameters of the parabola
The given parabola is \( y^2 = 4x \). We can compare it with the standard form \( y^2 = 4ax \) to find \( a \).

\[ a = 1 \]

### Step 2: Write the equation of the tangent to the parabola
The equation of the tangent to the parabola \( y^2 = 4x \) can be expressed as:

\[ y = mx + \frac{1}{m} \]

where \( m \) is the slope of the tangent.

### Step 3: Substitute the tangent equation into the hyperbola
For the hyperbola \( xy = 2 \), we substitute \( y \) from the tangent equation into the hyperbola equation:

\[ x\left(mx + \frac{1}{m}\right) = 2 \]

### Step 4: Simplify the equation
Expanding this gives:

\[ mx^2 + \frac{x}{m} - 2 = 0 \]

### Step 5: Use the condition for tangency
Since the line is a tangent to the hyperbola, the discriminant of this quadratic equation must be zero. For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Here, \( a = m \), \( b = \frac{1}{m} \), and \( c = -2 \). Thus, we set the discriminant to zero:

\[ \left(\frac{1}{m}\right)^2 - 4(m)(-2) = 0 \]

### Step 6: Solve the discriminant equation
This simplifies to:

\[ \frac{1}{m^2} + 8m = 0 \]

Multiplying through by \( m^2 \) (assuming \( m \neq 0 \)) gives:

\[ 1 + 8m^3 = 0 \]

Thus,

\[ 8m^3 = -1 \implies m^3 = -\frac{1}{8} \implies m = -\frac{1}{2} \]

### Step 7: Find the equation of the tangent
Now substituting \( m = -\frac{1}{2} \) back into the tangent equation:

\[ y = -\frac{1}{2}x - 2 \]

To express this in standard form, we multiply through by 2:

\[ 2y = -x - 4 \implies x + 2y + 4 = 0 \]

### Final Answer
The equation of the common tangent to the parabola \( y^2 = 4x \) and the hyperbola \( xy = 2 \) is:

\[ \boxed{x + 2y + 4 = 0} \]