Equation of the hyperbola with eccentricity `3/2` and foci at `(±2,0)` is

Options

A. `(x^(2))/4-(y^(2))/5=4/9`
B. `(x^(2))/4-(y^(2))/9=4/9`
C. `(x^(2))/4-(y^(2))/9=1`
D. none of these

Explanation

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

### Step 1: Identify the coordinates of the foci
The foci of the hyperbola are given as \( (±2, 0) \). This means that the distance from the center to each focus, denoted as \( c \), is \( c = 2 \).

### Step 2: Use the relationship between eccentricity, \( a \), and \( c \)
The eccentricity \( e \) of a hyperbola is defined as:
\[ e = \frac{c}{a} \]
Given that \( e = \frac{3}{2} \) and \( c = 2 \), we can set up the equation:
\[ \frac{2}{a} = \frac{3}{2} \]

### Step 3: Solve for \( a \)
Cross-multiplying gives:
\[ 2 \cdot 2 = 3 \cdot a \implies 4 = 3a \implies a = \frac{4}{3} \]

### Step 4: Find \( b^2 \) using the relationship \( c^2 = a^2 + b^2 \)
We know the relationship for hyperbolas:
\[ c^2 = a^2 + b^2 \]
Calculating \( a^2 \):
\[ a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
Now, substituting \( c = 2 \) into the equation gives:
\[ 2^2 = \frac{16}{9} + b^2 \implies 4 = \frac{16}{9} + b^2 \]
To solve for \( b^2 \), we first convert \( 4 \) to a fraction with a denominator of \( 9 \):
\[ 4 = \frac{36}{9} \]
Now substituting:
\[ \frac{36}{9} = \frac{16}{9} + b^2 \implies b^2 = \frac{36}{9} - \frac{16}{9} = \frac{20}{9} \]

### Step 5: Write the standard equation of the hyperbola
The standard form of the equation for a hyperbola centered at the origin with horizontal transverse axis is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = \frac{16}{9} \) and \( b^2 = \frac{20}{9} \):
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]
Multiplying through by \( 9 \) to eliminate the denominators:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 9 \]
Rearranging gives:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \]
This can be simplified to:
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]

### Step 6: Final form of the equation
Thus, the equation of the hyperbola is:
\[ \frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9} \]