Equation of one of the latusrectum of the hyperbola `(10x - 5)^2 + (10y - 2)^2 = 9(3x + 4y - 7)^2` is

Options

A. `y - 1//5 = - 3//4 (x - 1//2)`
B. `x - 1//5 = - 3//4 (y - 1//2)`
C. `y + 1//5 = - 3//4 (x + 1//2)`
D. `x + 1//5 = - 3//4 (y + 1//2)`

Answer

To find the equation of one of the latus rectum of the hyperbola given by the equation \((10x - 5)^2 + (10y - 2)^2 = 9(3x + 4y - 7)^2\), we will follow these steps:

### Step 1: Rewrite the given equation
The given equation is:
\[ (10x - 5)^2 + (10y - 2)^2 = 9(3x + 4y - 7)^2 \]
We can rewrite this equation in a more standard form by factoring out the constants.

### Step 2: Factor out constants
We can factor out \(10\) from the left-hand side:
\[ 10^2 \left( \left( x - \frac{1}{2} \right)^2 + \left( y - \frac{1}{5} \right)^2 \right) = 9(3x + 4y - 7)^2 \]
This simplifies to:
\[ 100 \left( \left( x - \frac{1}{2} \right)^2 + \left( y - \frac{1}{5} \right)^2 \right) = 9(3x + 4y - 7)^2 \]

### Step 3: Identify the center, focus, and directrix
From the equation, we can identify:
- The center of the hyperbola is at \(\left( \frac{1}{2}, \frac{1}{5} \right)\).
- The directrix is given by the line \(3x + 4y - 7 = 0\).

### Step 4: Find the eccentricity
The eccentricity \(e\) can be found from the equation. The left-hand side represents the distance from the center to a point on the hyperbola, while the right-hand side represents the distance to the directrix. The eccentricity can be calculated as:
\[ e = \frac{3}{2} \]

### Step 5: Equation of the latus rectum
The latus rectum of a hyperbola is a line that passes through the focus and is parallel to the directrix. The equation of the latus rectum can be derived from the focus and the slope of the directrix.

The slope of the directrix \(3x + 4y - 7 = 0\) is:
\[ \text{slope} = -\frac{3}{4} \]
The latus rectum will be perpendicular to this slope, so its slope will be:
\[ \text{slope of latus rectum} = \frac{4}{3} \]

### Step 6: Write the equation of the latus rectum
Using the point-slope form of the line equation, we can write the equation of the latus rectum:
\[ y - \frac{1}{5} = \frac{4}{3}\left(x - \frac{1}{2}\right) \]

### Step 7: Rearranging the equation
Rearranging this gives:
\[ y - \frac{1}{5} = \frac{4}{3}x - \frac{2}{3} \]
\[ y = \frac{4}{3}x + \left(\frac{1}{5} - \frac{2}{3}\right) \]
Calculating \(\frac{1}{5} - \frac{2}{3} = \frac{3 - 10}{15} = -\frac{7}{15}\):
\[ y = \frac{4}{3}x - \frac{7}{15} \]

### Final Answer
Thus, the equation of one of the latus rectum of the hyperbola is:
\[ \boxed{y = \frac{4}{3}x - \frac{7}{15}} \]