The shortest distance between the curves `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 and 4x^(2)+4y^(2)=a^(2)(bgta)` is f(a, b), then the value of `f(4, 6)+f(2, 3)` is

Answer

To solve the problem of finding the shortest distance between the curves given by the equations \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (a hyperbola) and \(4x^2 + 4y^2 = a^2\) (a circle), we will follow these steps:

### Step 1: Identify the curves
The first curve is a hyperbola given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The second curve is a circle, which can be rewritten as:
\[ x^2 + y^2 = \frac{a^2}{4} \]
This indicates that the radius of the circle is \(\frac{a}{2}\).

### Step 2: Determine the vertices of the hyperbola
The vertices of the hyperbola are located at \((a, 0)\) and \((-a, 0)\) on the x-axis.

### Step 3: Determine the center of the circle
The center of the circle is at the origin \((0, 0)\) with a radius of \(\frac{a}{2}\). This means the circle extends from \(-\frac{a}{2}\) to \(\frac{a}{2}\) along the x-axis.

### Step 4: Find the shortest distance between the curves
The shortest distance between the hyperbola and the circle occurs along the x-axis. The distance from the origin to the vertex of the hyperbola is \(a\), and the distance from the origin to the edge of the circle is \(\frac{a}{2}\).

Thus, the shortest distance \(D\) can be calculated as:
\[ D = a - \frac{a}{2} = \frac{a}{2} \]

### Step 5: Define the function \(f(a, b)\)
From the above calculation, we can define the function \(f(a, b)\) as:
\[ f(a, b) = \frac{a}{2} \]
Note that \(b\) does not affect the distance.

### Step 6: Calculate \(f(4, 6)\) and \(f(2, 3)\)
Now we will compute \(f(4, 6)\) and \(f(2, 3)\):
1. For \(f(4, 6)\):
\[ f(4, 6) = \frac{4}{2} = 2 \]
2. For \(f(2, 3)\):
\[ f(2, 3) = \frac{2}{2} = 1 \]

### Step 7: Find the final result
Now, we sum the two results:
\[ f(4, 6) + f(2, 3) = 2 + 1 = 3 \]

### Final Answer
Thus, the value of \(f(4, 6) + f(2, 3)\) is:
\[ \boxed{3} \]