Two common tangents to the circle `x^(2) + y^(2) = (a^(2))/(2)` and the parabola `y^(2) = 4ax` are

Options

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

Answer

To find the two common tangents to the circle given by the equation \( x^2 + y^2 = \frac{a^2}{2} \) and the parabola given by the equation \( y^2 = 4ax \), we will follow these steps:

### Step 1: Write the equation of the tangent to the circle
The equation of the tangent to the circle in slope form is given by:
\[ y = mx \pm \sqrt{\frac{a^2}{2}(1 + m^2)} \]
Here, \( r = \frac{a}{\sqrt{2}} \) is the radius of the circle.

### Step 2: Write the equation of the tangent to the parabola
The equation of the tangent to the parabola in slope form is given by:
\[ y = mx + \frac{a}{m} \]

### Step 3: Set the constant terms equal
For the tangents to be common, the constant terms from both equations must be equal. Thus, we set:
\[ \pm \sqrt{\frac{a^2}{2}(1 + m^2)} = \frac{a}{m} \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ \frac{a^2}{2}(1 + m^2) = \frac{a^2}{m^2} \]

### Step 5: Simplify the equation
Cancel \( a^2 \) from both sides (assuming \( a \neq 0 \)):
\[ \frac{1}{2}(1 + m^2) = \frac{1}{m^2} \]
Cross-multiplying gives:
\[ m^2(1 + m^2) = 2 \]

### Step 6: Rearrange the equation
Rearranging yields:
\[ m^4 + m^2 - 2 = 0 \]

### Step 7: Let \( t = m^2 \)
Let \( t = m^2 \), then we have:
\[ t^2 + t - 2 = 0 \]

### Step 8: Factor the quadratic
Factoring gives:
\[ (t - 1)(t + 2) = 0 \]
Thus, the solutions are:
\[ t = 1 \quad \text{or} \quad t = -2 \]
Since \( t = m^2 \), we discard \( t = -2 \) as it is not possible.

### Step 9: Solve for \( m \)
From \( t = 1 \), we have:
\[ m^2 = 1 \implies m = \pm 1 \]

### Step 10: Write the equations of the tangents
Substituting \( m = 1 \) and \( m = -1 \) back into the tangent equations:
1. For \( m = 1 \):
\[ y = x + \frac{a}{1} = x + a \]
2. For \( m = -1 \):
\[ y = -x + \frac{a}{-1} = -x - a \]

### Final Tangent Equations
Thus, the two common tangents are:
\[ y = x + a \quad \text{and} \quad y = -x - a \]