Find the equation of the hyperbola with focus at `(pm 2, 0) and e = (3)/(2) ` .

Answer

To find the equation of the hyperbola with given focus at \((\pm 2, 0)\) and eccentricity \(e = \frac{3}{2}\), we can follow these steps:

### Step 1: Identify the standard form of the hyperbola
The standard form of a hyperbola centered at the origin is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(2a\) is the distance between the vertices, and the foci are located at \((\pm ae, 0)\).

### Step 2: Determine the value of \(a\)
From the information given, the foci are at \((\pm 2, 0)\). This means:
\[ ae = 2 \]
Given that \(e = \frac{3}{2}\), we can substitute this into the equation:
\[ a \cdot \frac{3}{2} = 2 \]
To find \(a\), we can solve for \(a\):
\[ a = \frac{2}{\frac{3}{2}} = \frac{2 \cdot 2}{3} = \frac{4}{3} \]

### Step 3: Calculate \(b\) using the eccentricity formula
The relationship between \(a\), \(b\), and \(e\) is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the known values:
\[ \frac{3}{2} = \sqrt{1 + \frac{b^2}{\left(\frac{4}{3}\right)^2}} \]
Squaring both sides:
\[ \left(\frac{3}{2}\right)^2 = 1 + \frac{b^2}{\frac{16}{9}} \]
This simplifies to:
\[ \frac{9}{4} = 1 + \frac{9b^2}{16} \]
Subtracting 1 from both sides:
\[ \frac{9}{4} - 1 = \frac{9b^2}{16} \]
Converting 1 to a fraction with a denominator of 4:
\[ \frac{9}{4} - \frac{4}{4} = \frac{9b^2}{16} \]
This gives:
\[ \frac{5}{4} = \frac{9b^2}{16} \]
Cross-multiplying:
\[ 5 \cdot 16 = 4 \cdot 9b^2 \]
This simplifies to:
\[ 80 = 36b^2 \]
Dividing both sides by 36:
\[ b^2 = \frac{80}{36} = \frac{20}{9} \]

### Step 4: Write the equation of the hyperbola
Now that we have \(a^2\) and \(b^2\):
\[ a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
\[ b^2 = \frac{20}{9} \]
Substituting these values into the standard form of the hyperbola:
\[ \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \]
Multiplying through by 9 to eliminate the denominators:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \]
Thus, the final equation of the hyperbola is:
\[ \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \]