The foci of an ellipse (-5,-3) and (-5,-7) and the length of each latus rectum is 6. Find its equation

Given that the foci of the ellipse are \((-5, -3)\) and \((-5, -7)\), we can infer that the major axis is vertical because the x-coordinates of both foci are the same. To find the equation of the ellipse, let's proceed step by step.

### Step 1: Find the center
The center of the ellipse lies at the midpoint of the line segment connecting the foci. The midpoint is given by the formula:

\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

For the foci \((-5, -3)\) and \((-5, -7)\), the center is:

\[
\text{Center} = \left( \frac{-5 + (-5)}{2}, \frac{-3 + (-7)}{2} \right) = (-5, -5)
\]

Thus, the center of the ellipse is \((-5, -5)\).

### Step 2: Find the distance between the foci
The distance between the foci, denoted as \(2c\), is the vertical distance between \((-5, -3)\) and \((-5, -7)\). The formula for the distance between two points is:

\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Since the x-coordinates are the same, we only need the difference in the y-coordinates:

\[
2c = | -3 - (-7) | = | -3 + 7 | = 4
\]

Thus, \(c = 2\).

### Step 3: Use the length of the latus rectum
The length of each latus rectum is given as 6. The formula for the length of the latus rectum in an ellipse is:

\[
\text{Latus Rectum} = \frac{2b^2}{a}
\]

Given that the latus rectum length is 6, we have:

\[
6 = \frac{2b^2}{a}
\]

Thus,

\[
b^2 = 3a
\]

### Step 4: Use the relationship between \(a\), \(b\), and \(c\)
For an ellipse, the relationship between \(a\), \(b\), and \(c\) is:

\[
a^2 = b^2 + c^2
\]

Substitute \(b^2 = 3a\) and \(c = 2\) into the equation:

\[
a^2 = 3a + 2^2
\]

\[
a^2 = 3a + 4
\]

Rearranging this quadratic equation:

\[
a^2 - 3a - 4 = 0
\]

### Step 5: Solve for \(a\)
Solve the quadratic equation using the quadratic formula:

\[
a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}
\]

\[
a = \frac{3 \pm \sqrt{9 + 16}}{2}
\]

\[
a = \frac{3 \pm \sqrt{25}}{2}
\]

\[
a = \frac{3 \pm 5}{2}
\]

So,

\[
a = \frac{3 + 5}{2} = 4 \quad \text{(since \(a > 0\))}
\]

### Step 6: Find \(b^2\)
Now that we know \(a = 4\), we can find \(b^2\) using the earlier equation \(b^2 = 3a\):

\[
b^2 = 3(4) = 12
\]

### Step 7: Write the equation of the ellipse
Since the center of the ellipse is \((-5, -5)\), the general equation of the ellipse with a vertical major axis is:

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

Substituting \(h = -5\), \(k = -5\), \(a^2 = 16\), and \(b^2 = 12\), we get the equation:

\[
\frac{(x + 5)^2}{12} + \frac{(y + 5)^2}{16} = 1
\]

This is the equation of the ellipse.

---

Would you like more details on any of these steps or a deeper explanation of any concepts?

Here are some related questions you might find interesting:
1. How do you determine if an ellipse is vertical or horizontal?
2. What is the geometric significance of the latus rectum in an ellipse?
3. How can the relationship \(a^2 = b^2 + c^2\) be derived for ellipses?
4. How would the equation change if the foci were horizontal?
5. What are the similarities and differences between ellipses and hyperbolas?

**Tip:** Always check if an ellipse is vertical or horizontal based on the positions of the foci or the major axis!

The foci of an ellipse (-5,-3) and (-5,-7) and the length of each latus rectum is 6. Find its equation.

Learn how to find the equation of an ellipse with foci (-5, -3) and (-5, -7) and a latus rectum of 6. This solution covers key concepts in analytic geometry, including the midpoint formula, distance formula, and the relationship between a, b, and c in ellipses. Suitable for Grades 10-12.

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