The equation of common tangent of the curve `x^(2) + 4y^(2) = 8` and `y^(2) =4x` are

Options

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

Answer

To find the equations of the common tangents to the curves \( x^2 + 4y^2 = 8 \) (an ellipse) and \( y^2 = 4x \) (a parabola), we will follow these steps:

### Step 1: Identify the equations of the curves
The given curves are:
1. Ellipse: \( x^2 + 4y^2 = 8 \)
2. Parabola: \( y^2 = 4x \)

### Step 2: Write the equation of the tangent to the parabola
The standard form of the tangent to the parabola \( y^2 = 4ax \) is given by:
\[ y = mx + \frac{a}{m} \]
For the parabola \( y^2 = 4x \), we have \( a = 1 \). Thus, the equation of the tangent becomes:
\[ y = mx + \frac{1}{m} \]
This is Equation (1).

### Step 3: Write the equation of the tangent to the ellipse
The standard form of the tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by:
\[ y = mx + \sqrt{a^2m^2 + b^2} \]
For the ellipse \( x^2 + 4y^2 = 8 \), we can rewrite it in standard form:
\[ \frac{x^2}{8} + \frac{y^2}{2} = 1 \]
Here, \( a^2 = 8 \) and \( b^2 = 2 \), so \( a = \sqrt{8} = 2\sqrt{2} \) and \( b = \sqrt{2} \). The equation of the tangent becomes:
\[ y = mx + \sqrt{8m^2 + 2} \]
This is Equation (2).

### Step 4: Set the two tangent equations equal
To find the common tangents, we set Equation (1) equal to Equation (2):
\[ mx + \frac{1}{m} = mx + \sqrt{8m^2 + 2} \]
By canceling \( mx \) from both sides, we get:
\[ \frac{1}{m} = \sqrt{8m^2 + 2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{1}{m^2} = 8m^2 + 2 \]
Multiplying through by \( m^2 \) (assuming \( m \neq 0 \)):
\[ 1 = 8m^4 + 2m^2 \]
Rearranging gives:
\[ 8m^4 + 2m^2 - 1 = 0 \]

### Step 6: Substitute \( p = m^2 \)
Let \( p = m^2 \). Then the equation becomes:
\[ 8p^2 + 2p - 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ p = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ p = \frac{-2 \pm \sqrt{4 + 32}}{16} \]
\[ p = \frac{-2 \pm \sqrt{36}}{16} \]
\[ p = \frac{-2 \pm 6}{16} \]
This gives us two solutions:
1. \( p = \frac{4}{16} = \frac{1}{4} \)
2. \( p = \frac{-8}{16} = -\frac{1}{2} \) (not valid since \( p = m^2 \) cannot be negative)

Thus, \( m^2 = \frac{1}{4} \) implies \( m = \pm \frac{1}{2} \).

### Step 8: Find the equations of the tangents
For \( m = \frac{1}{2} \):
\[ y = \frac{1}{2}x + 2 \]
Multiplying through by 2 gives:
\[ x - 2y + 4 = 0 \]

For \( m = -\frac{1}{2} \):
\[ y = -\frac{1}{2}x - 2 \]
Multiplying through by 2 gives:
\[ x + 2y + 4 = 0 \]

### Final Answer
The equations of the common tangents are:
1. \( x - 2y + 4 = 0 \)
2. \( x + 2y + 4 = 0 \)