if `5x+9=0` is the directrix of the hyperbola `16x^(2)-9y^(2)=144`, then its corresponding focus is

Options

A. `(5,0)`
B. `(-(5)/(3),0)`
C. `(-5,0)`
D. `((5)/(3),0)`

Explanation

To solve the problem, we need to find the focus of the hyperbola given its equation and the directrix. Let's break it down step by step.

### Step 1: Write the equation of the hyperbola in standard form
The given equation of the hyperbola is:
\[ 16x^2 - 9y^2 = 144 \]

To convert it to the standard form, we divide the entire equation by 144:
\[ \frac{16x^2}{144} - \frac{9y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]
From this, we can identify \( a^2 = 9 \) and \( b^2 = 16 \).

### Step 2: Calculate \( a \) and \( b \)
Taking the square roots, we find:
\[ a = 3, \quad b = 4 \]

### Step 3: Calculate the eccentricity \( e \)
The eccentricity \( e \) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \]

### Step 4: Identify the directrix
The equation of the directrix for a hyperbola is given by:
\[ x = \frac{-a}{e} \]
We are given that the directrix is \( 5x + 9 = 0 \), which can be rearranged to:
\[ x = -\frac{9}{5} \]

### Step 5: Set the directrix equal to the formula
Now we equate the directrix to the formula:
\[ -\frac{a}{e} = -\frac{9}{5} \]
Substituting \( e = \frac{5}{3} \):
\[ -\frac{a}{\frac{5}{3}} = -\frac{9}{5} \]
This simplifies to:
\[ \frac{3a}{5} = \frac{9}{5} \]
Multiplying both sides by 5:
\[ 3a = 9 \]
Dividing by 3:
\[ a = 3 \]

### Step 6: Calculate the focus
The foci of a hyperbola are located at \( (\pm ae, 0) \). Therefore, we calculate:
\[ ae = 3 \cdot \frac{5}{3} = 5 \]
Thus, the coordinates of the foci are:
\[ (-5, 0) \quad \text{and} \quad (5, 0) \]

### Final Answer
The corresponding focus of the hyperbola is:
\[ \boxed{(-5, 0)} \]