Find the axes, vertices , foci , exentricity, equations of the directrices and length of the latus rectum of the hyperbola `5y ^(2) - 9x ^(2) = 36.`

Answer

To solve the problem step by step, we need to analyze the given hyperbola equation and extract the required information.

### Given Hyperbola Equation:
\[ 5y^2 - 9x^2 = 36 \]

### Step 1: Rewrite the Equation in Standard Form
We start by dividing the entire equation by 36 to put it in standard form.

\[ \frac{5y^2}{36} - \frac{9x^2}{36} = 1 \]

This simplifies to:

\[ \frac{y^2}{\frac{36}{5}} - \frac{x^2}{4} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), we identify:

\[ a^2 = \frac{36}{5}, \quad b^2 = 4 \]

### Step 3: Calculate \(a\) and \(b\)
Taking square roots, we find:

\[ a = \sqrt{\frac{36}{5}} = \frac{6}{\sqrt{5}}, \quad b = \sqrt{4} = 2 \]

### Step 4: Calculate \(c\) (the distance from the center to the foci)
Using the relationship \(c^2 = a^2 + b^2\):

\[ c^2 = \frac{36}{5} + 4 = \frac{36}{5} + \frac{20}{5} = \frac{56}{5} \]

Thus,

\[ c = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}} \]

### Step 5: Identify the Axes
Since the hyperbola is in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the transverse axis is along the y-axis.

### Step 6: Find the Vertices
The vertices are located at \((0, \pm a)\):

\[ \text{Vertices: } (0, \frac{6}{\sqrt{5}}) \text{ and } (0, -\frac{6}{\sqrt{5}}) \]

### Step 7: Find the Foci
The foci are located at \((0, \pm c)\):

\[ \text{Foci: } (0, \frac{2\sqrt{14}}{\sqrt{5}}) \text{ and } (0, -\frac{2\sqrt{14}}{\sqrt{5}}) \]

### Step 8: Calculate the Eccentricity
Eccentricity \(e\) is given by:

\[ e = \frac{c}{a} = \frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3} \]

### Step 9: Find the Equations of the Directrices
The equations of the directrices are given by \(y = \pm \frac{a}{e}\):

\[ \text{Directrix: } y = \pm \frac{\frac{6}{\sqrt{5}}}{\frac{\sqrt{14}}{3}} = \pm \frac{18}{\sqrt{5}\sqrt{14}} = \pm \frac{18}{\sqrt{70}} \]

### Step 10: Calculate the Length of the Latus Rectum
The length of the latus rectum \(L\) is given by:

\[ L = \frac{2b^2}{a} = \frac{2 \cdot 4}{\frac{6}{\sqrt{5}}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3} \]

### Summary of Results:
- **Axes:** Transverse axis along the y-axis
- **Vertices:** \((0, \frac{6}{\sqrt{5}})\) and \((0, -\frac{6}{\sqrt{5}})\)
- **Foci:** \((0, \frac{2\sqrt{14}}{\sqrt{5}})\) and \((0, -\frac{2\sqrt{14}}{\sqrt{5}})\)
- **Eccentricity:** \(e = \frac{\sqrt{14}}{3}\)
- **Equations of the Directrices:** \(y = \pm \frac{18}{\sqrt{70}}\)
- **Length of the Latus Rectum:** \(L = \frac{4\sqrt{5}}{3}\)