Find the number of points at which line `6y + 4x = 5` cuts hyperbola `((x - 1)^(2))/9 - ((y - 2)^(2))/4 = 1` .

Answer

To find the number of points at which the line \(6y + 4x = 5\) cuts the hyperbola \(\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{4} = 1\), we will follow these steps:

### Step 1: Rewrite the line equation
First, we rewrite the line equation in terms of \(y\):
\[ 6y + 4x = 5 \implies 6y = 5 - 4x \implies y = \frac{5 - 4x}{6} \]

### Step 2: Substitute \(y\) in the hyperbola equation
Next, we substitute \(y\) from the line equation into the hyperbola equation:
\[ \frac{(x - 1)^2}{9} - \frac{\left(\frac{5 - 4x}{6} - 2\right)^2}{4} = 1 \]
Simplifying the term inside the square:
\[ \frac{5 - 4x}{6} - 2 = \frac{5 - 4x - 12}{6} = \frac{-4x - 7}{6} \]
Now substituting this back into the hyperbola equation:
\[ \frac{(x - 1)^2}{9} - \frac{\left(\frac{-4x - 7}{6}\right)^2}{4} = 1 \]

### Step 3: Simplify the hyperbola equation
Now we can simplify the equation:
\[ \frac{(x - 1)^2}{9} - \frac{(16x^2 + 56x + 49)/36}{4} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{9} - \frac{16x^2 + 56x + 49}{144} = 1 \]

### Step 4: Clear the fractions
To eliminate the fractions, we can multiply through by \(144\):
\[ 16(x - 1)^2 - (16x^2 + 56x + 49) = 144 \]
Expanding \(16(x - 1)^2\):
\[ 16(x^2 - 2x + 1) = 16x^2 - 32x + 16 \]
Now substituting this in:
\[ 16x^2 - 32x + 16 - 16x^2 - 56x - 49 = 144 \]
This simplifies to:
\[ -32x - 56x + 16 - 49 = 144 \implies -88x - 33 = 144 \]

### Step 5: Solve for \(x\)
Rearranging gives:
\[ -88x = 144 + 33 \implies -88x = 177 \implies x = -\frac{177}{88} \]

### Step 6: Find \(y\)
Now substitute \(x\) back into the equation for \(y\):
\[ y = \frac{5 - 4\left(-\frac{177}{88}\right)}{6} = \frac{5 + \frac{708}{88}}{6} \]
This simplifies to:
\[ y = \frac{\frac{440 + 708}{88}}{6} = \frac{\frac{1148}{88}}{6} = \frac{1148}{528} = \frac{287}{132} \]

### Step 7: Conclusion
Thus, we find that the line intersects the hyperbola at exactly one point, which is:
\[ \left(-\frac{177}{88}, \frac{287}{132}\right) \]
Therefore, the number of points at which the line cuts the hyperbola is:
\[ \boxed{1} \]