find the common tangents of the circle `x^2+y^2=2a^2` and the parabola` y^2=8ax`

Options

A. `x = pm (y + 2a)`
B. `x = pm (y + a)`
C. `y = pm (x + 2a)`
D. `y = pm (x + a)`

Answer

To find the common tangents of the circle \( x^2 + y^2 = 2a^2 \) and the parabola \( y^2 = 8ax \), we can follow these steps:

### Step 1: Identify the equations of the conics
The given equations are:
- Circle: \( x^2 + y^2 = 2a^2 \)
- Parabola: \( y^2 = 8ax \)

### 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:
\[ y = mx + \frac{2a}{m} \]
In our case, since \( y^2 = 8ax \), we can rewrite it as:
\[ y^2 = 4(2a)x \]
Thus, the equation of the tangent to the parabola is:
\[ y = mx + \frac{4a}{m} \]

### Step 3: Find the distance from the center of the circle to the tangent line
The center of the circle is at the origin \( (0, 0) \), and the radius of the circle is \( \sqrt{2a^2} = \sqrt{2}a \).

The distance \( d \) from the point \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For the tangent line \( y = mx + \frac{4a}{m} \), we can rewrite it in the form \( mx - y + \frac{4a}{m} = 0 \). Here, \( A = m \), \( B = -1 \), and \( C = \frac{4a}{m} \).

Thus, the distance from the origin to the tangent line is:
\[ d = \frac{\left|\frac{4a}{m}\right|}{\sqrt{m^2 + 1}} \]

### Step 4: Set the distance equal to the radius
We set the distance equal to the radius of the circle:
\[ \frac{\left|\frac{4a}{m}\right|}{\sqrt{m^2 + 1}} = \sqrt{2}a \]
Removing \( a \) from both sides (assuming \( a \neq 0 \)):
\[ \frac{4}{|m| \sqrt{m^2 + 1}} = \sqrt{2} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4 = \sqrt{2} |m| \sqrt{m^2 + 1} \]
Squaring both sides results in:
\[ 16 = 2m^2(m^2 + 1) \]
This simplifies to:
\[ 16 = 2m^4 + 2m^2 \]
Dividing by 2:
\[ 8 = m^4 + m^2 \]
Rearranging gives:
\[ m^4 + m^2 - 8 = 0 \]

### Step 6: Let \( u = m^2 \)
Let \( u = m^2 \), then we have:
\[ u^2 + u - 8 = 0 \]
Using the quadratic formula:
\[ u = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2} \]

### Step 7: Solve for \( m \)
Calculating the values of \( u \):
1. \( u_1 = \frac{-1 + \sqrt{33}}{2} \)
2. \( u_2 = \frac{-1 - \sqrt{33}}{2} \) (not valid since \( u \) must be non-negative)

Thus:
\[ m^2 = \frac{-1 + \sqrt{33}}{2} \]
Taking the square root gives:
\[ m = \pm \sqrt{\frac{-1 + \sqrt{33}}{2}} \]

### Step 8: Write the equations of the tangents
Substituting \( m \) back into the tangent equation:
\[ y = mx + \frac{4a}{m} \]
We can derive two equations for the tangents corresponding to \( m \) and \( -m \).

### Final Tangent Equations
1. For \( m = \sqrt{\frac{-1 + \sqrt{33}}{2}} \):
\[ y = \sqrt{\frac{-1 + \sqrt{33}}{2}} x + \frac{4a}{\sqrt{\frac{-1 + \sqrt{33}}{2}}} \]

2. For \( m = -\sqrt{\frac{-1 + \sqrt{33}}{2}} \):
\[ y = -\sqrt{\frac{-1 + \sqrt{33}}{2}} x - \frac{4a}{\sqrt{\frac{-1 + \sqrt{33}}{2}}} \]