The center of a rectangular hyperbola lies on the line `y=2xdot` If one of the asymptotes is `x+y+c=0` , then the other asymptote is `6x+3y-4c=0` (b) `3x+6y-5c=0` `3x-6y-c=0` (d) none of these

Options

A. `6x+3y-4c=0`
B. `3x+6y-5c=0`
C. `3x-6y-c=0`
D. none of these

Explanation

To solve the problem step by step, we need to find the other asymptote of the rectangular hyperbola given that one of the asymptotes is \( x + y + c = 0 \) and that the center of the hyperbola lies on the line \( y = 2x \).

### Step 1: Understand the Given Information
We know that:
- The center of the hyperbola lies on the line \( y = 2x \).
- One asymptote is given by the equation \( x + y + c = 0 \).

### Step 2: Rewrite the Line Equation
The equation \( y = 2x \) can be rewritten in standard form:
\[ 2x - y = 0 \]

### Step 3: Determine the Slope of the Given Asymptote
The slope of the line \( x + y + c = 0 \) can be found by rewriting it in slope-intercept form:
\[ y = -x - c \]
Thus, the slope of this asymptote is \( -1 \).

### Step 4: Find the Slope of the Other Asymptote
Since the asymptotes of a rectangular hyperbola are perpendicular to each other, the product of their slopes must equal \(-1\). Let the slope of the other asymptote be \( m \):
\[ (-1) \cdot m = -1 \implies m = 1 \]
Thus, the slope of the other asymptote is \( 1 \).

### Step 5: Write the Equation of the Other Asymptote
The equation of a line with slope \( 1 \) that passes through the center (which lies on \( y = 2x \)) can be expressed as:
\[ y - 2x = \lambda \quad \text{(where \( \lambda \) is a constant)} \]
Rearranging gives:
\[ y - 2x - \lambda = 0 \]

### Step 6: Use the Condition of Concurrency
Since the center lies on both asymptotes, we can set up a determinant to find \( \lambda \). The lines are concurrent if the determinant of the coefficients is zero. The equations we have are:
1. \( 2x - y = 0 \) (from \( y = 2x \))
2. \( x + y + c = 0 \)
3. \( y - 2x - \lambda = 0 \)

The determinant is:
\[ \begin{vmatrix} 2 & -1 & 0 \\ 1 & 1 & c \\ -2 & 1 & -\lambda \end{vmatrix} = 0 \]

### Step 7: Calculate the Determinant
Calculating the determinant:
\[ = 2(1)(-\lambda) + 1(1)(0) + 0(-1)(-2) - 0(1)(-2) - (-1)(1)(-\lambda) - (-1)(2)(c) \]
This simplifies to:
\[ -2\lambda + \lambda + 2c = 0 \implies -\lambda + 2c = 0 \implies \lambda = 2c \]

### Step 8: Substitute Back to Find the Other Asymptote
Substituting \( \lambda = 2c \) back into the equation of the other asymptote:
\[ y - 2x - 2c = 0 \implies 2x - y + 2c = 0 \]

### Step 9: Rearranging the Equation
Rearranging gives:
\[ 2x - y = -2c \implies 2x - y + 2c = 0 \]

### Conclusion
Thus, the equation of the other asymptote can be expressed as:
\[ 2x - y + 2c = 0 \]
This does not match any of the provided options directly. However, if we multiply through by a factor to match one of the options, we can see that it can be manipulated into the form of one of the choices.

### Final Answer
The other asymptote is:
\[ \text{(d) none of these} \]