Equation of the hyperbola with eccentricity `3/2` and foci at `(pm 2, 0)` is `5x^(2) - 4y^(2) = k^(2)`. What is the value of `K`?

Options

A. `4//3`
B. `3//4`
C. `(4//3sqrt5)`
D. `(3//4sqrt5)`

Answer

To find the value of \( k \) for the hyperbola with eccentricity \( \frac{3}{2} \) and foci at \( (\pm 2, 0) \), we can follow these steps:

### Step 1: Understand the Hyperbola's Properties
The standard form of the hyperbola centered at the origin with a horizontal transverse axis is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( 2c \) is the distance between the foci, \( c \) is the distance from the center to each focus, and \( e \) is the eccentricity defined as \( e = \frac{c}{a} \).

### Step 2: Identify the Values of \( c \) and \( a \)
From the given foci \( (\pm 2, 0) \), we have:
\[ c = 2 \]
The eccentricity is given as:
\[ e = \frac{3}{2} \]
Using the relationship \( e = \frac{c}{a} \), we can find \( a \):
\[ \frac{3}{2} = \frac{2}{a} \implies a = \frac{2 \cdot 2}{3} = \frac{4}{3} \]

### Step 3: Find \( a^2 \)
Now, we calculate \( a^2 \):
\[ a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]

### Step 4: Relate \( b^2 \) to \( a^2 \) and \( c^2 \)
We know that for hyperbolas:
\[ c^2 = a^2 + b^2 \]
Substituting the known values:
\[ 2^2 = \frac{16}{9} + b^2 \implies 4 = \frac{16}{9} + b^2 \]
To solve for \( b^2 \), convert \( 4 \) to a fraction with a denominator of \( 9 \):
\[ 4 = \frac{36}{9} \]
Now, substituting:
\[ \frac{36}{9} = \frac{16}{9} + b^2 \]
Subtract \( \frac{16}{9} \) from both sides:
\[ b^2 = \frac{36}{9} - \frac{16}{9} = \frac{20}{9} \]

### Step 5: Write the Equation of the Hyperbola
The equation of the hyperbola can now be written as:
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]
Multiplying through by \( 9 \) gives:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \]
Rearranging gives:
\[ \frac{5x^2}{4} - \frac{4y^2}{5} = 1 \]

### Step 6: Compare with Given Equation
The given equation is:
\[ 5x^2 - 4y^2 = k^2 \]
To match the forms, we can multiply the entire equation by \( k^2 \):
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = \frac{k^2}{9} \]
Thus, we can equate:
\[ k^2 = 9 \]

### Conclusion
Taking the square root gives:
\[ k = 3 \]

### Final Answer
The value of \( k \) is \( 3 \).