The equations of the directrices of `16x^(2) + 25y^(2) - 96x - 100y - 156 = 0` are

Options

A. `9(x-3)=pm25`
B. `3(x-3)=pm16`
C. `3(y-2)=pm25`
D. `3(x-3)=pm25`

Answer

To find the equations of the directrices of the ellipse given by the equation \(16x^2 + 25y^2 - 96x - 100y - 156 = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 16x^2 + 25y^2 - 96x - 100y - 156 = 0 \]
Rearranging gives:
\[ 16x^2 - 96x + 25y^2 - 100y = 156 \]

### Step 2: Completing the Square for \(x\) and \(y\)
For the \(x\) terms:
\[ 16(x^2 - 6x) \quad \text{(factor out 16)} \]
To complete the square, take half of \(-6\) (which is \(-3\)), square it to get \(9\), and rewrite:
\[ 16((x - 3)^2 - 9) = 16(x - 3)^2 - 144 \]

For the \(y\) terms:
\[ 25(y^2 - 4y) \quad \text{(factor out 25)} \]
Take half of \(-4\) (which is \(-2\)), square it to get \(4\), and rewrite:
\[ 25((y - 2)^2 - 4) = 25(y - 2)^2 - 100 \]

### Step 3: Substitute Back into the Equation
Substituting back into the rearranged equation:
\[ 16((x - 3)^2 - 9) + 25((y - 2)^2 - 4) = 156 \]
This simplifies to:
\[ 16(x - 3)^2 - 144 + 25(y - 2)^2 - 100 = 156 \]
Combine the constants:
\[ 16(x - 3)^2 + 25(y - 2)^2 - 244 = 156 \]
\[ 16(x - 3)^2 + 25(y - 2)^2 = 400 \]

### Step 4: Divide by 400
Now, divide the entire equation by 400:
\[ \frac{16(x - 3)^2}{400} + \frac{25(y - 2)^2}{400} = 1 \]
This simplifies to:
\[ \frac{(x - 3)^2}{25} + \frac{(y - 2)^2}{16} = 1 \]

### Step 5: Identify Parameters of the Ellipse
From the standard form of the ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
We have:
- \(h = 3\)
- \(k = 2\)
- \(a^2 = 25\) (thus \(a = 5\))
- \(b^2 = 16\) (thus \(b = 4\))

### Step 6: Calculate the Eccentricity
The eccentricity \(e\) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 7: Find the Equations of the Directrices
The equations of the directrices for an ellipse are given by:
\[ x = h \pm \frac{a}{e} \]
Substituting the values:
\[ x = 3 \pm \frac{5}{\frac{3}{5}} = 3 \pm \frac{25}{3} \]
Thus, we have:
\[ x = 3 + \frac{25}{3} \quad \text{and} \quad x = 3 - \frac{25}{3} \]
Calculating these:
1. \(x = 3 + \frac{25}{3} = \frac{9}{3} + \frac{25}{3} = \frac{34}{3}\)
2. \(x = 3 - \frac{25}{3} = \frac{9}{3} - \frac{25}{3} = -\frac{16}{3}\)

### Final Answer
The equations of the directrices are:
\[ x = \frac{34}{3} \quad \text{and} \quad x = -\frac{16}{3} \]