The slope of one of the common tangent to circle `x^(2)+y^(2)=1` and ellipse `x^(2)/4+2y^(2)=1` is `sqrt(a/b)` where `gcd(a, b)=1` then `(a+b)/2` is equal to

Answer

To find the slope of one of the common tangents to the circle \(x^2 + y^2 = 1\) and the ellipse \(\frac{x^2}{4} + 2y^2 = 1\), we can follow these steps:

### Step 1: Identify the equations of the tangent lines
For the circle \(x^2 + y^2 = 1\), the equation of the tangent line in slope-intercept form is given by:
\[ y = mx \pm \sqrt{1 + m^2} \]
where \(r = 1\) for the circle.

For the ellipse \(\frac{x^2}{4} + 2y^2 = 1\), we rewrite it in standard form:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a^2 = 4\) and \(b^2 = \frac{1}{2}\). The equation of the tangent line in slope-intercept form is:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \(a^2\) and \(b^2\):
\[ y = mx \pm \sqrt{4m^2 + \frac{1}{2}} \]

### Step 2: Set the equations equal for common tangents
Since we are looking for common tangents, we set the two equations equal to each other:
\[ mx \pm \sqrt{1 + m^2} = mx \pm \sqrt{4m^2 + \frac{1}{2}} \]
We can cancel \(mx\) from both sides:
\[ \sqrt{1 + m^2} = \sqrt{4m^2 + \frac{1}{2}} \]

### Step 3: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ 1 + m^2 = 4m^2 + \frac{1}{2} \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ 1 - \frac{1}{2} = 4m^2 - m^2 \]
\[ \frac{1}{2} = 3m^2 \]

### Step 5: Solve for \(m^2\)
From the equation \(\frac{1}{2} = 3m^2\), we can solve for \(m^2\):
\[ m^2 = \frac{1}{6} \]

### Step 6: Find \(m\)
Taking the square root gives:
\[ m = \pm \sqrt{\frac{1}{6}} = \pm \frac{1}{\sqrt{6}} = \pm \frac{\sqrt{6}}{6} \]

### Step 7: Express \(m\) in the form \(\sqrt{\frac{a}{b}}\)
We can express \(m\) as:
\[ m = \frac{1}{\sqrt{6}} = \sqrt{\frac{1}{6}} \]
Here, \(a = 1\) and \(b = 6\).

### Step 8: Calculate \((a + b)/2\)
Now, we need to find \((a + b)/2\):
\[ \frac{a + b}{2} = \frac{1 + 6}{2} = \frac{7}{2} = 3.5 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{3.5} \]