Math  /  Geometry

QuestionFind the standard form of the equation of the el

Studdy Solution

STEP 1

What is this asking? We need to find the equation of an ellipse centered at the origin, knowing how far it stretches along the x and y axes. Watch out! Don't mix up the a and b values!
They're tied to the x and y stretches, and putting them in the wrong spot messes things up.

STEP 2

1. Identify key features from the graph.
2. Plug values into the standard ellipse equation.

STEP 3

Alright, let's **do this**!
We're given a graph of an ellipse centered at (0,0) (0,0) , which is super helpful.
This tells us our equation will be in the nice, simple form: x2a2+y2b2=1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

STEP 4

Now, peep that horizontal stretch!
The ellipse goes **6 units** to the left and right of the center.
This horizontal stretch is our *a* value, so a=6 a = 6 .

STEP 5

Next up, the vertical action!
The ellipse stretches **4 units** up and down from the center.
This vertical stretch is our *b* value, meaning b=4 b = 4 .

STEP 6

Remember our standard equation? x2a2+y2b2=1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 Let's **plug in** our *a* and *b* values.

STEP 7

Substituting a=6 a = 6 , we get a2=62=36 a^2 = 6^2 = 36 .
Substituting b=4 b = 4 , we find b2=42=16 b^2 = 4^2 = 16 .

STEP 8

**Putting it all together**, our equation becomes: x236+y216=1 \frac{x^2}{36} + \frac{y^2}{16} = 1 Boom!

STEP 9

The standard form of the ellipse's equation is x236+y216=1 \frac{x^2}{36} + \frac{y^2}{16} = 1 .

Was this helpful?

Studdy solves anything!

banner

Start learning now

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

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord