Equation of the hyperbola with foci (+-2 ,0) and eccentricity 3/2 is

Options

A. `x^2/9-y^(2)/4=-1`
B. `x^(2)/(20)-y^(2)/(16)=-1`
C. `x^(2)/4-y^(2)/5=4/9`
D. `(9x^(2))/(20)-(9y^(2))/(16)=-1`

Answer

To find the equation of the hyperbola with given foci and eccentricity, we can follow these steps:

### Step 1: Identify the given parameters
The foci of the hyperbola are given as \((\pm 2, 0)\) and the eccentricity \(e\) is given as \(\frac{3}{2}\).

### Step 2: Determine the value of \(c\)
The distance from the center to each focus is denoted as \(c\). Since the foci are at \((\pm 2, 0)\), we have:
\[ c = 2 \]

### Step 3: Use the relationship between eccentricity, \(a\), and \(c\)
The eccentricity \(e\) is related to \(a\) (the distance from the center to a vertex) and \(c\) by the formula:
\[ e = \frac{c}{a} \]
Substituting the known values:
\[ \frac{3}{2} = \frac{2}{a} \]
Cross-multiplying gives:
\[ 3a = 4 \implies a = \frac{4}{3} \]

### Step 4: Find \(b^2\) using the relationship \(c^2 = a^2 + b^2\)
We can find \(b^2\) using the formula:
\[ c^2 = a^2 + b^2 \]
Calculating \(c^2\) and \(a^2\):
\[ c^2 = 2^2 = 4 \]
\[ a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
Now substituting into the equation:
\[ 4 = \frac{16}{9} + b^2 \]
To isolate \(b^2\), we first convert 4 into a fraction with a denominator of 9:
\[ 4 = \frac{36}{9} \]
Now substituting:
\[ \frac{36}{9} = \frac{16}{9} + b^2 \]
Subtracting \(\frac{16}{9}\) from both sides:
\[ b^2 = \frac{36}{9} - \frac{16}{9} = \frac{20}{9} \]

### Step 5: Write the equation of the hyperbola
The standard form of the equation of a hyperbola centered at the origin with a horizontal transverse axis is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \(a^2\) and \(b^2\):
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]
This can be rewritten as:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \]

### Step 6: Simplify the equation
Multiplying through by 9 gives:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \]
This can be rearranged to:
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]
This is the final equation of the hyperbola.

### Final Answer
The equation of the hyperbola is:
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]