The eccentricity of the hyperbola ` 9x^(2) -16y^(2) =144 ` is

Options

A. `4//3`
B. `5//4`
C. `4//5`
D. `3//5`

Answer

To find the eccentricity of the hyperbola given by the equation \( 9x^2 - 16y^2 = 144 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 9x^2 - 16y^2 = 144 \]
To convert this into standard form, we divide every term by 144:
\[ \frac{9x^2}{144} - \frac{16y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = 16 \quad \text{and} \quad b^2 = 9 \]
Thus, we find:
\[ a = \sqrt{16} = 4 \quad \text{and} \quad b = \sqrt{9} = 3 \]

### Step 3: Calculate \(c\)
The value of \(c\) (the distance from the center to the foci) is given by the formula:
\[ c = \sqrt{a^2 + b^2} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ c = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 4: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \frac{c}{a} \]
Substituting the values we found:
\[ e = \frac{5}{4} \]

### Conclusion
Thus, the eccentricity of the hyperbola \( 9x^2 - 16y^2 = 144 \) is:
\[ \boxed{\frac{5}{4}} \]

---