An ellipse is represented by the equation [tex]\(\frac{(x-5)^2}{625}+\frac{(y-4)^2}{225}=1\)[/tex].

Each directrix of this ellipse is a:

A. horizontal line that is 20 units from the center on the major axis.
B. vertical line that is 20 units from the center on the major axis.
C. horizontal line that is 31.25 units from the center on the major axis.
D. vertical line that is 31.25 units from the center on the major axis.



Answer :

Sure! Let's start solving the problem step-by-step. The given equation of the ellipse is:

[tex]\[ \frac{(x-5)^2}{625} + \frac{(y-4)^2}{225} = 1. \][/tex]

### Step 1: Identify the Parameters of the Ellipse
- Center [tex]\((h, k)\)[/tex]: [tex]\((5, 4)\)[/tex]
- Semi-major axis [tex]\(a\)[/tex]: The denominator under the [tex]\(x\)[/tex] term [tex]\(\sqrt{625} = 25\)[/tex]
- Semi-minor axis [tex]\(b\)[/tex]: The denominator under the [tex]\(y\)[/tex] term [tex]\(\sqrt{225} = 15\)[/tex]

### Step 2: Determine Which Axis is the Major Axis
Since [tex]\(625 > 225\)[/tex], the semi-major axis [tex]\(a = 25\)[/tex] lies along the [tex]\(x\)[/tex]-axis. Therefore, the ellipse is oriented horizontally.

### Step 3: Calculate the Focal Distance
The focal distance [tex]\(c\)[/tex] is found using the relationship:
[tex]\[ c = \sqrt{a^2 - b^2} \][/tex]
where [tex]\(a = 25\)[/tex] and [tex]\(b = 15\)[/tex].
[tex]\[ c = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20 \][/tex]

### Step 4: Determine the Directrix Distance
For horizontal ellipses, the distance of the directrix from the center is calculated using:
[tex]\[ \text{Directrix distance} = \frac{a^2}{c} \][/tex]
where [tex]\(a = 25\)[/tex] and [tex]\(c = 20\)[/tex].
[tex]\[ \text{Directrix distance} = \frac{25^2}{20} = \frac{625}{20} = 31.25 \][/tex]

### Step 5: Interpret the Result
Since the major axis is horizontal, the directrix lines are vertical and located 31.25 units away from the center on each side along the [tex]\(x\)[/tex]-axis.

So the correct answer is:

[tex]\[ \boxed{\text{vertical line that is 31.25 units}} \][/tex]