Math / Geometry

Question- Use the imformation provided to writo the stand From equation of each ellipse.
Uertices: $(6,-6)(-10,-6)$ Foci $(-2+2 \sqrt{7,-6})(-2-2 \sqrt{7},-6)$ .

Studdy Solution

STEP 1

What is this asking? We need to find the equation of an ellipse, given its vertices and foci. Watch out! Remember the difference between the $a$ , $b$ , and $c$ values, and where they go in the ellipse equation!
Also, make sure you center the ellipse correctly.
Don't just assume the center is at $(0, 0)$ !

STEP 2

1. Find the Center
2. Find a
3. Find c
4. Find b
5. Write the Equation

STEP 3

The center of the ellipse is the midpoint of the vertices.
Let's calculate it!
The midpoint formula is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ .

STEP 4

Our vertices are $(6, -6)$ and $(-10, -6)$ , so plugging those into the midpoint formula gives us $\left(\frac{6 + (-10)}{2}, \frac{-6 + (-6)}{2}\right)$ .

STEP 5

Simplifying, we get $\left(\frac{-4}{2}, \frac{-12}{2}\right)$ , which gives us the center at $(-2, -6)$ .
Awesome!

STEP 6

The value of $a$ is the distance from the center to a vertex.
We can use the distance formula to find this distance.
Remember, the distance formula is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

STEP 7

Let's use the center $(-2, -6)$ and one of the vertices, $(6, -6)$ .
Plugging these into the distance formula gives us $\sqrt{(6 - (-2))^2 + (-6 - (-6))^2}$ .

STEP 8

Simplifying, we get $\sqrt{8^2 + 0^2} = \sqrt{64} = 8$ .
So, $a = \mathbf{8}$ !

STEP 9

The value of $c$ is the distance from the center to a focus.
We'll use the distance formula again.

STEP 10

We have the center $(-2, -6)$ and a focus $(-2 + 2\sqrt{7}, -6)$ .
Plugging these into the distance formula gives us $\sqrt{((-2 + 2\sqrt{7}) - (-2))^2 + (-6 - (-6))^2}$ .

STEP 11

Simplifying, we get $\sqrt{(2\sqrt{7})^2 + 0^2} = \sqrt{4 \cdot 7} = \sqrt{28} = 2\sqrt{7}$ .
So, $c = \mathbf{2\sqrt{7}}$ .

STEP 12

Now, we need to find $b$ .
We know the relationship between $a$ , $b$ , and $c$ in an ellipse is $c^2 = a^2 - b^2$ .
We can rearrange this to $b^2 = a^2 - c^2$ .

STEP 13

We know $a = 8$ and $c = 2\sqrt{7}$ , so $b^2 = 8^2 - (2\sqrt{7})^2 = 64 - 28 = 36$ .
Taking the square root of both sides gives us $b = \mathbf{6}$ .

STEP 14

Since the vertices and foci have the same $y$ -coordinate, the major axis is horizontal.
The general equation for an ellipse with a horizontal major axis and center $(h, k)$ is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ .

STEP 15

We found the center to be $(-2, -6)$ , so $h = -2$ and $k = -6$ .
We also found $a = 8$ and $b = 6$ .
Plugging these values into the equation, we get $\frac{(x - (-2))^2}{8^2} + \frac{(y - (-6))^2}{6^2} = 1$ .

STEP 16

Simplifying, we get our final equation: $\frac{(x + 2)^2}{64} + \frac{(y + 6)^2}{36} = 1$ .

STEP 17

The standard form equation of the ellipse is $\frac{(x + 2)^2}{64} + \frac{(y + 6)^2}{36} = 1$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms