For the equation of rectangular hyperbola `xy = 18`

Options

A. Length of transverse axis = length of conjugate axis = 12
B. Vertices are `(3sqrt(2),3sqrt(2))` or `(-3sqrt(2), -3sqrt(2))`
C. Foci are `( 6 + 6), (-6, -6)`
D. Equation of tangent with slope 1 cannot be possible

Answer

To solve the problem regarding the rectangular hyperbola given by the equation \(xy = 18\), we will follow these steps:

### Step 1: Identify the standard form of the hyperbola
The equation of a rectangular hyperbola can be expressed in the form \(xy = c^2\). Here, we have \(c^2 = 18\).

### Step 2: Calculate the value of \(c\)
To find \(c\), we take the square root of \(c^2\):
\[ c = \sqrt{18} = 3\sqrt{2} \]

### Step 3: Determine the lengths of the transverse and conjugate axes
For a rectangular hyperbola, the lengths of the transverse axis and the conjugate axis are given by:
\[ \text{Length of transverse axis} = 2c \]
\[ \text{Length of conjugate axis} = 2c \]
Substituting \(c = 3\sqrt{2}\):
\[ \text{Length of transverse axis} = 2(3\sqrt{2}) = 6\sqrt{2} \]
\[ \text{Length of conjugate axis} = 2(3\sqrt{2}) = 6\sqrt{2} \]

### Step 4: Calculate the coordinates of the vertices
The vertices of a rectangular hyperbola are located at \((c, c)\) and \((-c, -c)\):
\[ \text{Vertices} = (3\sqrt{2}, 3\sqrt{2}) \text{ and } (-3\sqrt{2}, -3\sqrt{2}) \]

### Step 5: Determine the coordinates of the foci
The foci of a rectangular hyperbola are located at \((c\sqrt{2}, c\sqrt{2})\) and \((-c\sqrt{2}, -c\sqrt{2})\):
\[ \text{Foci} = (6, 6) \text{ and } (-6, -6) \]

### Step 6: Analyze the tangent line with a given slope
For a rectangular hyperbola, the slope of the tangent line at a point \((ct, \frac{c}{t})\) is given by:
\[ \text{slope} = -\frac{1}{t^2} \]
If we set the slope equal to 1:
\[ 1 = -\frac{1}{t^2} \]
This leads to \(t^2 = -1\), which is not possible since \(t^2\) cannot be negative.

### Summary of Results
1. Length of transverse axis = 12
2. Length of conjugate axis = 12
3. Vertices = \((3\sqrt{2}, 3\sqrt{2})\) and \((-3\sqrt{2}, -3\sqrt{2})\)
4. Foci = \((6, 6)\) and \((-6, -6)\)
5. Slope of tangent with slope 1 is not possible.