What is the equation to the hyperbola if its latusrectum is `9/2` and eccentricity is `5/4`.

Answer

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

### Step 1: Use the formula for the length of the latus rectum
The length of the latus rectum (L) of a hyperbola is given by the formula:
\[ L = \frac{2b^2}{a} \]
Given that the latus rectum is \( \frac{9}{2} \), we can set up the equation:
\[ \frac{2b^2}{a} = \frac{9}{2} \]

### Step 2: Solve for \( b^2 \) in terms of \( a \)
Multiplying both sides by \( a \) and rearranging gives:
\[ 2b^2 = \frac{9a}{2} \]
Dividing both sides by 2:
\[ b^2 = \frac{9a}{4} \]
This is our first equation.

### Step 3: Use the formula for eccentricity
The eccentricity (e) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Given that the eccentricity is \( \frac{5}{4} \), we can square both sides:
\[ \left(\frac{5}{4}\right)^2 = 1 + \frac{b^2}{a^2} \]
Calculating \( \left(\frac{5}{4}\right)^2 \):
\[ \frac{25}{16} = 1 + \frac{b^2}{a^2} \]

### Step 4: Rearranging the equation
Subtracting 1 from both sides:
\[ \frac{25}{16} - 1 = \frac{b^2}{a^2} \]
Converting 1 to a fraction with a denominator of 16:
\[ \frac{25}{16} - \frac{16}{16} = \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{9}{16} = \frac{b^2}{a^2} \]

### Step 5: Substitute \( b^2 \) from Step 2
From Step 2, we have \( b^2 = \frac{9a}{4} \). Substituting this into the equation:
\[ \frac{9}{16} = \frac{\frac{9a}{4}}{a^2} \]
This simplifies to:
\[ \frac{9}{16} = \frac{9}{4a} \]

### Step 6: Solve for \( a \)
Cross-multiplying gives:
\[ 9 \cdot 4 = 16 \cdot 9a \]
Cancelling out the 9:
\[ 36 = 16a \]
Dividing both sides by 16:
\[ a = \frac{36}{16} = \frac{9}{4} \]

### Step 7: Find \( b^2 \)
Now substituting \( a \) back into the equation for \( b^2 \):
\[ b^2 = \frac{9a}{4} = \frac{9 \cdot \frac{9}{4}}{4} = \frac{81}{16} \]

### Step 8: Write the equation of the hyperbola
The standard form of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = \left(\frac{9}{4}\right)^2 = \frac{81}{16} \) and \( b^2 = \frac{81}{16} \):
\[ \frac{x^2}{\frac{81}{16}} - \frac{y^2}{\frac{81}{16}} = 1 \]
This simplifies to:
\[ \frac{x^2}{\frac{81}{16}} - \frac{y^2}{\frac{81}{16}} = 1 \]
Multiplying through by \( \frac{81}{16} \):
\[ x^2 - y^2 = \frac{81}{16} \]

### Final Equation
Thus, the equation of the hyperbola is:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]