The locus of the point `( (e^(t) +e^(-t))/( 2),(e^t-e^(-t))/(2))` is a hyperbola of eccentricity

Options

A. `sqrt3`
B. 3
C. `sqrt2`
D. 2

Answer

To find the locus of the point \(\left( \frac{e^{t} + e^{-t}}{2}, \frac{e^{t} - e^{-t}}{2} \right)\) and determine the eccentricity of the hyperbola, we can follow these steps:

### Step 1: Define the coordinates
Let:
\[ h = \frac{e^{t} + e^{-t}}{2} \]
\[ k = \frac{e^{t} - e^{-t}}{2} \]

### Step 2: Express \(e^{t}\) and \(e^{-t}\) in terms of \(h\) and \(k\)
From the definitions of \(h\) and \(k\), we can express \(e^{t}\) and \(e^{-t}\):
1. From \(h\):
\[ e^{t} + e^{-t} = 2h \quad \text{(Equation 1)} \]
2. From \(k\):
\[ e^{t} - e^{-t} = 2k \quad \text{(Equation 2)} \]

### Step 3: Add and subtract the equations
Adding Equation 1 and Equation 2:
\[ (e^{t} + e^{-t}) + (e^{t} - e^{-t}) = 2h + 2k \]
This simplifies to:
\[ 2e^{t} = 2h + 2k \implies e^{t} = h + k \]

Subtracting Equation 2 from Equation 1:
\[ (e^{t} + e^{-t}) - (e^{t} - e^{-t}) = 2h - 2k \]
This simplifies to:
\[ 2e^{-t} = 2h - 2k \implies e^{-t} = h - k \]

### Step 4: Use the expressions for \(e^{t}\) and \(e^{-t}\)
Now we can express \(e^{t}\) and \(e^{-t}\) in terms of \(h\) and \(k\):
\[ e^{t} = h + k \]
\[ e^{-t} = h - k \]

### Step 5: Find the relationship between \(h\) and \(k\)
We can use the identity \(e^{t} e^{-t} = 1\):
\[ (h + k)(h - k) = 1 \]
Expanding this gives:
\[ h^2 - k^2 = 1 \]

### Step 6: Write the equation in standard form
Rearranging the equation:
\[ h^2 - k^2 = 1 \]
This is the standard form of a hyperbola:
\[ \frac{h^2}{1} - \frac{k^2}{1} = 1 \]

### Step 7: Identify \(a\) and \(b\)
From the equation, we can identify:
\[ a^2 = 1 \implies a = 1 \]
\[ b^2 = 1 \implies b = 1 \]

### Step 8: Calculate the eccentricity
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{a^2}{b^2}} \]
Substituting the values of \(a\) and \(b\):
\[ e = \sqrt{1 + \frac{1^2}{1^2}} = \sqrt{1 + 1} = \sqrt{2} \]

### Conclusion
The eccentricity of the hyperbola is \(\sqrt{2}\).

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