The equation of one of the latusrectum of `(x-3)^(2)/(5)+(y-5)^(2)/(9)=1` is

Options

A. y-3=0
B. x-1=0
C. y=6
D. x=5

Answer

To find the equation of one of the latus rectum of the ellipse given by the equation

\[ \frac{(x-3)^2}{5} + \frac{(y-5)^2}{9} = 1, \]

we will follow these steps:

### Step 1: Identify the center and the semi-major and semi-minor axes
The standard form of the ellipse is

\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \]

where \((h, k)\) is the center, \(a^2\) is the semi-major axis squared, and \(b^2\) is the semi-minor axis squared.

From the given equation:
- \(h = 3\)
- \(k = 5\)
- \(a^2 = 5\) (thus \(a = \sqrt{5}\))
- \(b^2 = 9\) (thus \(b = 3\))

### Step 2: Determine the orientation of the ellipse
Since \(b^2 > a^2\), the major axis is vertical. The latus rectum of a vertical ellipse is given by the equation:

\[ x = h \pm \frac{b^2}{a}. \]

### Step 3: Calculate the value of \(\frac{b^2}{a}\)
Now we compute:

\[ \frac{b^2}{a} = \frac{9}{\sqrt{5}}. \]

### Step 4: Write the equations of the latus rectum
Now we can find the equations of the latus rectum:

1. First latus rectum:
\[ x = 3 + \frac{9}{\sqrt{5}}. \]

2. Second latus rectum:
\[ x = 3 - \frac{9}{\sqrt{5}}. \]

### Step 5: Finalize the equations
Thus, the equations of the latus rectum are:

\[ x = 3 + \frac{9}{\sqrt{5}} \quad \text{and} \quad x = 3 - \frac{9}{\sqrt{5}}. \]