Rectangular hyperbola is the hyperbola whose asymptotes are perpendicular hence its equationis `x^(2) - y^(2) = a^(2)`, if axes are rotated by `45^(@)` in clockwise direction then its equation becomes `xy = c^(2)`.
Focus of hyperbola `xy = 16`, is

Options

A. `(4sqrt(2), 4sqrt(2))`
B. `(4sqrt(2),0)`
C. `(0, 4sqrt(2))`
D. `(4, 0)`

Answer

To find the focus of the hyperbola given by the equation \( xy = 16 \), we can follow these steps:

### Step 1: Understand the given hyperbola
The equation \( xy = c^2 \) represents a rectangular hyperbola. In this case, \( c^2 = 16 \), so \( c = 4 \).

### Step 2: Relate the hyperbola to its standard form
The standard form of a rectangular hyperbola is given by \( xy = c^2 \). The foci of a rectangular hyperbola can be derived from its parameters.

### Step 3: Find the value of \( a \)
For the hyperbola \( xy = c^2 \), we know that the relationship between \( c \) and \( a \) is given by:
\[ c^2 = 2a^2 \]
Substituting \( c^2 = 16 \):
\[ 16 = 2a^2 \]
Dividing both sides by 2:
\[ a^2 = 8 \implies a = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Determine the coordinates of the foci
The foci of the hyperbola \( xy = c^2 \) are located at \( (a\sqrt{2}, a\sqrt{2}) \) and \( (-a\sqrt{2}, -a\sqrt{2}) \). Substituting \( a = 2\sqrt{2} \):
\[ \text{Focus 1: } (2\sqrt{2}\sqrt{2}, 2\sqrt{2}\sqrt{2}) = (4, 4) \]
\[ \text{Focus 2: } (-2\sqrt{2}\sqrt{2}, -2\sqrt{2}\sqrt{2}) = (-4, -4) \]

### Step 5: Conclusion
Thus, the foci of the hyperbola \( xy = 16 \) are:
- \( (4, 4) \)
- \( (-4, -4) \)

### Final Answer
The focus of the hyperbola \( xy = 16 \) is \( (4\sqrt{2}, 4\sqrt{2}) \) and \( (-4\sqrt{2}, -4\sqrt{2}) \).

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