Equations of the latus recta of the ellipse `9x^(2)+4y^(2)-18x-8y-23=0` are

Options

A. `y=pmsqrt(5)`
B. `x=pmsqrt(5)`
C. `y=1 pmsqrt(5)`
D. `x=-1 pmsqrt(5)`

Answer

To find the equations of the latus recta of the given ellipse \(9x^2 + 4y^2 - 18x - 8y - 23 = 0\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 9x^2 + 4y^2 - 18x - 8y - 23 = 0 \]
Rearranging gives:
\[ 9x^2 - 18x + 4y^2 - 8y - 23 = 0 \]

### Step 2: Complete the square for \(x\) and \(y\)
For \(x\):
\[ 9(x^2 - 2x) \quad \text{(factor out 9)} \]
Completing the square:
\[ x^2 - 2x = (x - 1)^2 - 1 \implies 9[(x - 1)^2 - 1] = 9(x - 1)^2 - 9 \]

For \(y\):
\[ 4(y^2 - 2y) \quad \text{(factor out 4)} \]
Completing the square:
\[ y^2 - 2y = (y - 1)^2 - 1 \implies 4[(y - 1)^2 - 1] = 4(y - 1)^2 - 4 \]

### Step 3: Substitute back into the equation
Substituting back gives:
\[ 9((x - 1)^2 - 1) + 4((y - 1)^2 - 1) - 23 = 0 \]
Simplifying:
\[ 9(x - 1)^2 - 9 + 4(y - 1)^2 - 4 - 23 = 0 \]
\[ 9(x - 1)^2 + 4(y - 1)^2 - 36 = 0 \]
\[ 9(x - 1)^2 + 4(y - 1)^2 = 36 \]

### Step 4: Divide by 36 to get the standard form
\[ \frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{9} = 1 \]

### Step 5: Identify \(a\), \(b\), \(h\), and \(k\)
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\):
- \(h = 1\)
- \(k = 1\)
- \(a^2 = 4 \implies a = 2\)
- \(b^2 = 9 \implies b = 3\)

### Step 6: Find the equations of the latus recta
The equations of the latus recta for an ellipse are given by:
\[ y = k \pm \frac{b^2}{a} \]
Calculating \(\frac{b^2}{a}\):
\[ \frac{b^2}{a} = \frac{9}{2} = 4.5 \]
Thus, the equations of the latus recta are:
\[ y = 1 + 4.5 \quad \text{and} \quad y = 1 - 4.5 \]
This simplifies to:
\[ y = 5.5 \quad \text{and} \quad y = -3.5 \]

### Final Answer
The equations of the latus recta are:
\[ y = 5.5 \quad \text{and} \quad y = -3.5 \]