Center at (2,-5), conjugate axis parallel to the y-axis, slopes at asymptotes numerically one-sixteenth times the length of the latus rectum, and distance between foci is 2sqrt145.

Somehow, somewhere, there is a question associated with this.

To find the equation of an ellipse with the given properties, we'll break down the problem into steps.

Step 1: Identify the center of the ellipse.
The given information states that the center of the ellipse is at (2,-5).

Step 2: Determine the orientation of the ellipse.
Since the conjugate axis is parallel to the y-axis, we know that this is a vertical ellipse.

Step 3: Find the length of the latus rectum.
The length of the latus rectum can be calculated using the formula:
L = 2b^2/a, where a represents the semi-major axis and b represents the semi-minor axis.

In this case, since the ellipse is vertical, a = b. Therefore, the formula simplifies to:
L = 2b^2/b = 2b.

We're also given that the slopes of the asymptotes are numerically one-sixteenth times the length of the latus rectum. So, the slopes of the asymptotes are 1/16 * L.

Step 4: Find the equation of the asymptotes.
For a vertical ellipse, the equation of the asymptotes is given by:
y = ±(x - h) * m + k,
where (h, k) is the center of the ellipse, and m is the slope of the asymptotes.

Using the information given, we can write the equations of the asymptotes as:
y = ±(x - 2) * (1/16 * L) - 5.

Step 5: Find the semi-major axis, the semi-minor axis, and the eccentricity.
To find the semi-major axis (a), we divide the length of the latus rectum by 2:
a = L/2 = 2b/2 = b.

To find the semi-minor axis (b), we're given that a = b, so we can take a = b = L/2.

To find the eccentricity (e), we use the formula:
e = c/a,
where c is the distance between the foci. We're given that the distance between the foci is 2√145, so c = 2√145.

Now, we can find e:
e = (2√145)/(L/2) = 4√145/L.

Step 6: Write the equation of the ellipse.
The equation of a vertical ellipse with center (h, k) = (2, -5), semi-major axis (a) = b = L/2, and eccentricity (e) is given by:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1.

Substituting the known values, we get:
(x - 2)^2 / (L/2)^2 + (y + 5)^2 / (L/2)^2 = 1.

So, the equation of the ellipse with the given properties is:
(x - 2)^2 / (b)^2 + (y + 5)^2 / (b)^2 = 1,
where b = L/2.