Equation of common tangent of parabola `y ^(2) = 8x and x ^(2) + y =0` is

Options

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

Answer

To find the equation of the common tangent to the parabolas \( y^2 = 8x \) and \( x^2 + y = 0 \), we will follow these steps:

### Step 1: Identify the parameters of the parabolas
The first parabola is given by \( y^2 = 8x \). This can be compared to the standard form \( y^2 = 4ax \):
- Here, \( 4a = 8 \) implies \( a = 2 \).

The second parabola is given by \( x^2 + y = 0 \), which can be rewritten as \( y = -x^2 \). This can be compared to the standard form \( x^2 = 4by \):
- Here, \( 4b = -1 \) implies \( b = -\frac{1}{4} \).

### Step 2: Write the equations of the tangents
For the first parabola \( y^2 = 8x \), the equation of the tangent line can be expressed as:
\[ y = mx + \frac{a}{m} = mx + \frac{2}{m} \]

For the second parabola \( x^2 + y = 0 \) (or \( y = -x^2 \)), the equation of the tangent line can be expressed as:
\[ y = mx - bm^2 = mx + \frac{1}{4}m^2 \]

### Step 3: Set the tangents equal to each other
Since we are looking for a common tangent, we set the two equations equal:
\[ mx + \frac{2}{m} = mx + \frac{1}{4}m^2 \]

### Step 4: Simplify the equation
Cancelling \( mx \) from both sides gives:
\[ \frac{2}{m} = \frac{1}{4}m^2 \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying yields:
\[ 2 \cdot 4 = m^3 \implies 8 = m^3 \]

### Step 6: Solve for \( m \)
Taking the cube root of both sides gives:
\[ m = 2 \]

### Step 7: Substitute \( m \) back into the tangent equation
Now, substituting \( m = 2 \) back into the tangent equation for the first parabola:
\[ y = 2x + \frac{2}{2} = 2x + 1 \]

### Step 8: Rearranging to standard form
Rearranging this equation gives:
\[ 2x - y + 1 = 0 \]

### Conclusion
Thus, the equation of the common tangent to the parabolas \( y^2 = 8x \) and \( x^2 + y = 0 \) is:
\[ \boxed{2x - y + 1 = 0} \]