Question

Find the equation of the ellipse which has vertices \(\mathrm{V}_{1}(-2,6), \mathrm{V}_{2}(-2,-4)\), and foci \(\mathrm{F}_{1}(-2,4), \mathrm{F}_{2}(-2,-2)\) (See figure.)

Short answer

The equation of the ellipse with vertices \(\mathrm{V}_{1}(-2,6),\mathrm{V}_{2}(-2,-4)\) and foci \(\mathrm{F}_{1}(-2,4), \mathrm{F}_{2}(-2,-2)\) is \(\frac{(x + 2)^2}{25} + \frac{(y - 1)^2}{16} = 1\).

Step-by-step solution

1. Find the center of the ellipse.

The center of the ellipse is the midpoint between the vertices or between the foci. Since we are given coordinates of the vertices \(\mathrm{V}_{1}(-2,6)\) and \(\mathrm{V}_{2}(-2,-4)\), let's find the midpoint. The midpoint formula is: Midpoint = \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\) Plugging in the coordinates of the vertices, we get: Midpoint, which is the center (\(h, k\)) = \(\left(\frac{-2 + (-2)}{2}, \frac{6 + (-4)}{2}\right) = (-2,1)\) So, the center of the ellipse is \((-2,1)\).

2. Determine the lengths of the semi-major and semi-minor axes.

The semi-major axis, denoted as \(a\), is the distance from the center of the ellipse to one of the vertices, and the semi-minor axis, denoted as \(b\), is the distance from the center of the ellipse to one of the co-vertices. The distance between the center and a vertex is the semi-major axis. Using the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) The distance from the center to vertex \(\mathrm{V}_{1}\) (semi-major axis, a) = \(\sqrt{((-2) - (-2))^2 + (6 - 1)^2} = \sqrt{25} = 5\) To find the semi-minor axis length, we can use the relationship between the distance from the center to one focus (which we'll call \(c\)), the semi-major axis length (which is \(a\)), and the semi-minor axis length (which we'll call \(b\)). The relationship is: \(c^2 = a^2 - b^2\) Since the coordinates of the foci are given as \(\mathrm{F}_{1}(-2,4)\) and \(\mathrm{F}_{2}(-2,-2)\), we can find the distance between the center and focus using the distance formula. The distance between the center and focus \(\mathrm{F}_{1}\) = \(\sqrt{((-2) - (-2))^2 + (4 - 1)^2} = \sqrt{9} = 3\) So, \(c = 3\). Now, solving for \(b\): \(3^2 = 5^2 - b^2 \Rightarrow b^2 = 25 - 9 \Rightarrow b^2 = 16 \Rightarrow b = 4\)

3. Write the equation of the ellipse.

Now that we have the center and the lengths of the semi-major and semi-minor axes, we can write the equation of the ellipse in standard form. The general equation of an ellipse centered at \((h,k)\) is: \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\) Plugging in the center and the values of \(a\) and \(b\), we get: \(\frac{(x - (-2))^2}{5^2} + \frac{(y - 1)^2}{4^2} = 1\) Simplifying, we get the equation of the ellipse: \(\frac{(x + 2)^2}{25} + \frac{(y - 1)^2}{16} = 1\) This is the equation of the ellipse with vertices \(\mathrm{V}_{1}(-2,6),\mathrm{V}_{2}(-2,-4)\) and foci \(\mathrm{F}_{1}(-2,4), \mathrm{F}_{2}(-2,-2)\).

Key concepts

Midpoint Formula

When dealing with geometry, especially ellipses, finding the midpoint is crucial as it often reveals the center.
The midpoint formula is a straightforward way to find the point exactly between two given points.
This is particularly useful when analyzing ellipses, as the center can be determined using vertices or foci locations.

The midpoint formula states:

• Given two points i.e., \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint \( (h, k) \) is: \(((x_1 + x_2)/2, (y_1 + y_2)/2)\).

For example, if you know the vertices \(\mathrm{V}_1(-2,6)\) and \(\mathrm{V}_2(-2,-4)\), simply plug them into the formula to find the midpoint which is the center: \((-2, 1)\).
Once you have the midpoint, you know where the ellipse is centered on the graph.

Distance Formula

The distance formula helps us find how far apart two points are in a Cartesian plane.
This is essential for locating the semi-major and semi-minor axes of an ellipse.
The formula comes into play to calculate the distances from the center to the vertices and foci.

The distance formula is provided by:

• For any two points \((x_1, y_1)\) and \((x_2, y_2)\), their distance \(d\) is:\(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

For an ellipse with a center at \((-2, 1)\), you can find the distance (semi-major axis) to a vertex such as \((-2, 6)\). Using the formula, this distance results in \(d = 5\).
This formula is versatile and can also be used to find distances to the foci.

Semi-Major Axis

In the realm of ellipses, the semi-major axis is crucial!
It is the longest radius stretching from the center to any vertex.
Basically, it's half of the longest diameter of the ellipse. Knowing this value makes constructing and analyzing the ellipse much easier.

The semi-major axis can be found once the center and a vertex are known.
Apply the distance formula between them.

• For instance, if the center is located at \((-2,1)\) and a vertex at \((-2,6)\), use the distance formula to get 5 as the semi-major axis length.

This denotes how far out the longest stretching point in the ellipse extends from its center.

Semi-Minor Axis

The semi-minor axis is the lesser-known sibling of the semi-major axis, but it is no less important.
It represents half the shortest diameter of the ellipse and runs perpendicular to the semi-major axis at the center.

While you can approach it similarly using physical measurements, the most efficient method is to use the relationship: \(c^2 = a^2 - b^2\), where:

• \(c\) is the distance from the center to a focus,
• \(a\) is the semi-major axis, and
• \(b\) is the semi-minor axis.

For example, in our case with \(a = 5\) and \(c = 3\), solve for \(b\) by substituting:\(b^2 = 25 - 9 => b = 4\).
The semi-minor axis is significant for understanding the distance across the ellipse in the narrower direction.

Standard Form of Ellipse Equation

An ellipse's equation tells you all about its shape and positioning, but it can seem a bit daunting at first.
The equation's purpose is to express an ellipse centered at any given point with designated lengths for its axes.

The standard form of an ellipse equation, centered at \((h, k)\), appears as:

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

In this format:

• \(a\) is the length of the semi-major axis, and
• \(b\) is the length of the semi-minor axis.

For example, given the calculated center \((-2,1)\), \(a=5\), and \(b=4\), this specific ellipse has: \(\frac{(x+2)^2}{25} + \frac{(y-1)^2}{16} = 1\).
This equation encapsulates and captures all relevant data about the ellipse once computed.