For the hyperbola `9x^(2)-16y^(2)-72x+96y-144=0`.
(a) Foci …………. (b) Eccentricity ………….
(c ) L.R. ………….. (d) Directrices …………….

Answer

To solve the hyperbola equation \(9x^2 - 16y^2 - 72x + 96y - 144 = 0\) and find the foci, eccentricity, length of the latus rectum (L.R.), and directrices, we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 9x^2 - 16y^2 - 72x + 96y - 144 = 0 \]
Rearranging gives:
\[ 9x^2 - 72x - 16y^2 + 96y = 144 \]

### Step 2: Completing the Square
Next, we complete the square for both \(x\) and \(y\).

**For \(x\):**
\[ 9(x^2 - 8x) = 9((x - 4)^2 - 16) = 9(x - 4)^2 - 144 \]

**For \(y\):**
\[ -16(y^2 - 6y) = -16((y - 3)^2 - 9) = -16(y - 3)^2 + 144 \]

Substituting back into the equation:
\[ 9(x - 4)^2 - 144 - 16(y - 3)^2 + 144 = 144 \]
This simplifies to:
\[ 9(x - 4)^2 - 16(y - 3)^2 = 144 \]

### Step 3: Dividing by 144
Divide the entire equation by 144 to get it into standard form:
\[ \frac{(x - 4)^2}{16} - \frac{(y - 3)^2}{9} = 1 \]
This is now in the standard form of a hyperbola:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
where \(h = 4\), \(k = 3\), \(a^2 = 16\) (thus \(a = 4\)), and \(b^2 = 9\) (thus \(b = 3\)).

### Step 4: Finding the Center
The center of the hyperbola is at the point \((h, k) = (4, 3)\).

### Step 5: Finding the Foci
The distance \(c\) from the center to the foci is given by:
\[ c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
Thus, the foci are located at:
\[ (4 \pm c, 3) = (4 \pm 5, 3) \Rightarrow (9, 3) \text{ and } (-1, 3) \]

### Step 6: Finding the Eccentricity
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \frac{c}{a} = \frac{5}{4} \]

### Step 7: Finding the Length of the Latus Rectum (L.R.)
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2} \]

### Step 8: Finding the Directrices
The equations of the directrices for a hyperbola are given by:
\[ x = h \pm \frac{a^2}{c} \]
Calculating:
\[ \frac{a^2}{c} = \frac{16}{5} \]
Thus, the directrices are:
\[ x = 4 \pm \frac{16}{5} \Rightarrow x = 4 + \frac{16}{5} = \frac{36}{5} \text{ and } x = 4 - \frac{16}{5} = \frac{-4}{5} \]

### Summary of Results
(a) Foci: \((9, 3)\) and \((-1, 3)\)
(b) Eccentricity: \(\frac{5}{4}\)
(c) Length of L.R.: \(\frac{9}{2}\)
(d) Directrices: \(x = \frac{36}{5}\) and \(x = \frac{-4}{5}\)