The equations of the latus recta of the ellipse `9x^(2) + 25y^(2) - 36x + 50y - 164 = 0` are

Options

A. x-4=0,x+2=0
B. x-6=0,x+2=0
C. x-6=0,x-2=0
D. x+4=0,x+5=0

Answer

To find the equations of the latus recta of the ellipse given by the equation \(9x^2 + 25y^2 - 36x + 50y - 164 = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ 9x^2 + 25y^2 - 36x + 50y - 164 = 0 \]
Rearranging it gives:
\[ 9x^2 - 36x + 25y^2 + 50y - 164 = 0 \]

### Step 2: Completing the Square for \(x\) and \(y\)
We will complete the square for the \(x\) and \(y\) terms.

**For \(x\):**
\[ 9(x^2 - 4x) = 9((x - 2)^2 - 4) = 9(x - 2)^2 - 36 \]

**For \(y\):**
\[ 25(y^2 + 2y) = 25((y + 1)^2 - 1) = 25(y + 1)^2 - 25 \]

Substituting these back into the equation gives:
\[ 9((x - 2)^2 - 4) + 25((y + 1)^2 - 1) - 164 = 0 \]
This simplifies to:
\[ 9(x - 2)^2 - 36 + 25(y + 1)^2 - 25 - 164 = 0 \]
Combining like terms:
\[ 9(x - 2)^2 + 25(y + 1)^2 - 225 = 0 \]
Thus, we have:
\[ 9(x - 2)^2 + 25(y + 1)^2 = 225 \]

### Step 3: Dividing by 225
Dividing the entire equation by 225:
\[ \frac{9(x - 2)^2}{225} + \frac{25(y + 1)^2}{225} = 1 \]
This simplifies to:
\[ \frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1 \]

### Step 4: Identifying Parameters
From the standard form of the ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
we identify:
- \(h = 2\)
- \(k = -1\)
- \(a^2 = 25 \Rightarrow a = 5\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 5: Finding the Eccentricity
The eccentricity \(e\) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 6: Equations of the Latus Recta
The equations of the latus recta are given by:
\[ x = h \pm ae \]
Substituting the values:
\[ x = 2 \pm 5 \cdot \frac{4}{5} \]
This simplifies to:
\[ x = 2 \pm 4 \]
Thus, we find:
\[ x = 6 \quad \text{or} \quad x = -2 \]

### Final Step: Writing the Equations
The equations of the latus recta are:
\[ x - 6 = 0 \quad \text{and} \quad x + 2 = 0 \]