Which of the following equations in parametric form can represent a hyperbola, where `t` is a parameter?

Options

A. `A: x=(a)/(2)(t+(1)/(t)) and y=(b)/(2)(t-(1)/(t))`
B. `B: (tx)/(a)-(y)/(b)+t=0 and (x)/(a)+(ty)/(b)-1=0`
C. `C: x=e^(t)+e^(-t) and y=e^(t)-e^(-t)`
D. `D: x^(2)-6=2cost and y^(2)+2=4cos^(2)((t)/(2))`

Answer

To determine which of the given parametric equations represents a hyperbola, we will analyze each option step by step.

### Step 1: Analyze Option A
The equations are:
- \( x = \frac{a}{2} T + \frac{1}{T} \)
- \( y = \frac{b}{2} T - \frac{1}{T} \)

**1.1** Rewrite \( x \) and \( y \) in terms of \( T \):
- \( \frac{x}{\frac{a}{2}} = T + \frac{2}{aT} \)
- \( \frac{y}{\frac{b}{2}} = T - \frac{2}{bT} \)

**1.2** Square both equations:
- \( \left(\frac{x}{\frac{a}{2}}\right)^2 = \left(T + \frac{2}{aT}\right)^2 \)
- \( \left(\frac{y}{\frac{b}{2}}\right)^2 = \left(T - \frac{2}{bT}\right)^2 \)

**1.3** Expand and simplify:
- \( \frac{x^2}{\frac{a^2}{4}} = T^2 + 2 + \frac{4}{a^2T^2} \)
- \( \frac{y^2}{\frac{b^2}{4}} = T^2 - 2 + \frac{4}{b^2T^2} \)

**1.4** Combine the equations:
- \( \frac{x^2}{\frac{a^2}{4}} - \frac{y^2}{\frac{b^2}{4}} = 1 \)

**Conclusion for A**: This represents a hyperbola.

### Step 2: Analyze Option B
The equations are:
- \( \frac{Tx}{a} - \frac{y}{b} + T = 0 \)
- \( \frac{x}{a} + \frac{Ty}{b} - 1 = 0 \)

**2.1** Rearranging gives:
- \( Tx = \frac{y}{b} - T \)
- \( x = a(1 - \frac{Ty}{b}) \)

**2.2** Substitute \( x \) into the first equation and simplify:
- This leads to a quadratic form that does not yield the hyperbola form.

**Conclusion for B**: This does not represent a hyperbola.

### Step 3: Analyze Option C
The equations are:
- \( x = e^T + e^{-T} \)
- \( y = e^T - e^{-T} \)

**3.1** Square both equations:
- \( x^2 = (e^T + e^{-T})^2 = e^{2T} + 2 + e^{-2T} \)
- \( y^2 = (e^T - e^{-T})^2 = e^{2T} - 2 + e^{-2T} \)

**3.2** Combine:
- \( x^2 - y^2 = 4 \)

**3.3** Rewrite:
- \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \)

**Conclusion for C**: This represents a hyperbola.

### Step 4: Analyze Option D
The equations are:
- \( x^2 - 6 = 2 \cos T \)
- \( y^2 + 2 = 4 \cos^2 \frac{T}{2} \)

**4.1** Rearranging gives:
- \( x^2 = 2 \cos T + 6 \)
- \( y^2 = 4 \cos^2 \frac{T}{2} - 2 \)

**4.2** Use trigonometric identities to combine:
- This leads to the form \( x^2 - y^2 = 6 \).

**4.3** Rewrite:
- \( \frac{x^2}{6} - \frac{y^2}{6} = 1 \)

**Conclusion for D**: This represents a hyperbola.

### Final Conclusion
The equations that represent a hyperbola are:
- Option A
- Option C
- Option D