the equations to the directrices of the ellipse `4(x-3)^(2)+9(y+2)^(2)=144,` are

Options

A. `5x-15+-18sqrt(5)=0`
B. `5x+15+-2sqrt(5)=0`
C. `15x+-2sqrt(5)=0`
D. `15x-5+-18sqrt(5)=0`

Answer

To find the equations of the directrices of the ellipse given by the equation \( 4(x-3)^2 + 9(y+2)^2 = 144 \), we will follow these steps:

### Step 1: Convert the equation to the standard form of the ellipse
We start by dividing the entire equation by 144 to express it in the standard form:

\[ \frac{4(x-3)^2}{144} + \frac{9(y+2)^2}{144} = 1 \]

This simplifies to:

\[ \frac{(x-3)^2}{36} + \frac{(y+2)^2}{16} = 1 \]

### Step 2: Identify the center and the values of \(a^2\) and \(b^2\)
From the standard form, we can identify:
- The center \(C\) of the ellipse is at \((3, -2)\).
- \(a^2 = 36\) and \(b^2 = 16\).

Thus, we have:
- \(a = \sqrt{36} = 6\)
- \(b = \sqrt{16} = 4\)

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the values of \(b^2\) and \(a^2\):

\[ e = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 4: Find the equations of the directrices
The equations of the directrices for an ellipse centered at the origin are given by:

\[ x = \pm \frac{a}{e} \]

Calculating \(\frac{a}{e}\):

\[ \frac{a}{e} = \frac{6}{\frac{\sqrt{5}}{3}} = 6 \cdot \frac{3}{\sqrt{5}} = \frac{18}{\sqrt{5}} \]

Thus, the equations of the directrices are:

\[ x = \pm \frac{18}{\sqrt{5}} \]

### Step 5: Adjust for the center of the ellipse
Since the center of the ellipse is at \((3, -2)\), we need to adjust the equations of the directrices by shifting them from the origin to the center:

\[ x - 3 = \pm \frac{18}{\sqrt{5}} \]

This gives us:

\[ x = 3 \pm \frac{18}{\sqrt{5}} \]

### Final Step: Write the final equations
Thus, the final equations of the directrices are:

\[ x = 3 + \frac{18}{\sqrt{5}} \quad \text{and} \quad x = 3 - \frac{18}{\sqrt{5}} \]

### Summary of the Solution
The equations of the directrices of the ellipse \(4(x-3)^2 + 9(y+2)^2 = 144\) are:

\[ 5x - 15 = \pm 18\sqrt{5} \]