The equation of common tangent to the parabola `y^2 =8x` and hyperbola `3x^2 -y^2=3` is

Options

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

Answer

To find the equation of the common tangent to the parabola \( y^2 = 8x \) and the hyperbola \( 3x^2 - y^2 = 3 \), we will follow these steps:

### Step 1: Write the equations in standard form
1. **Parabola**: The equation \( y^2 = 8x \) can be rewritten in the standard form \( y^2 = 4ax \), where \( 4a = 8 \). Thus, \( a = 2 \).
2. **Hyperbola**: The equation \( 3x^2 - y^2 = 3 \) can be rewritten as \( \frac{x^2}{1} - \frac{y^2}{3} = 1 \). Here, \( a^2 = 1 \) and \( b^2 = 3 \), so \( a = 1 \) and \( b = \sqrt{3} \).

### Step 2: Find the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
Substituting \( a^2 = 1 \) and \( b^2 = 3 \):
\[ y = mx \pm \sqrt{m^2 - 3} \]

### Step 3: Find the equation of the tangent to the parabola
The equation of the tangent to the parabola \( y^2 = 4ax \) is given by:
\[ y = mx + \frac{a}{m} \]
Substituting \( a = 2 \):
\[ y = mx + \frac{2}{m} \]

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

### Step 5: Solve for \( m \)
By simplifying, we can cancel \( mx \) from both sides:
\[ \pm \sqrt{m^2 - 3} = \frac{2}{m} \]
Squaring both sides gives:
\[ m^2 - 3 = \frac{4}{m^2} \]
Multiplying through by \( m^2 \) to eliminate the fraction:
\[ m^4 - 3m^2 - 4 = 0 \]
Let \( t = m^2 \), then we have:
\[ t^2 - 3t - 4 = 0 \]

### Step 6: Factor or use the quadratic formula
Factoring gives:
\[ (t - 4)(t + 1) = 0 \]
Thus, \( t = 4 \) or \( t = -1 \). Since \( t = m^2 \) cannot be negative, we take \( t = 4 \), which gives \( m^2 = 4 \) and \( m = \pm 2 \).

### Step 7: Find the equations of the tangents
1. For \( m = 2 \):
\[ y = 2x + \frac{2}{2} = 2x + 1 \quad \Rightarrow \quad 2x - y + 1 = 0 \]
2. For \( m = -2 \):
\[ y = -2x - 1 \quad \Rightarrow \quad 2x + y + 1 = 0 \]

### Final Answer
The equations of the common tangents are:
1. \( 2x - y + 1 = 0 \)
2. \( 2x + y + 1 = 0 \)