Math

QuestionDetermine the standard equation, center, and radius of the ellipse from 4x2+4y216=04x^{2}+4y^{2}-16=0.

Studdy Solution

STEP 1

Assumptions1. The given equation is of an ellipse. . The standard form of the equation of an ellipse is xa+yb=1\frac{x^{}}{a^{}} + \frac{y^{}}{b^{}} =1 where (0,0)(0,0) is the center of the ellipse and aa and bb are the lengths of the semi-major and semi-minor axes respectively.
3. The center of the ellipse is at the origin (0,0) as there are no xx or yy terms in the equation.
4. The radius of an ellipse is not a single value, but rather two values the lengths of the semi-major and semi-minor axes. However, in this case, since the coefficients of xx^{} and yy^{} are equal, the ellipse is actually a circle and the radius is a single value.

STEP 2

First, we need to rewrite the given equation in the standard form of an ellipse equation.
4x2+4y216=04x^{2}+4y^{2}-16=0

STEP 3

To do this, we first add16 to both sides of the equation to isolate the terms with x2x^{2} and y2y^{2} on one side.
x2+y2=16x^{2}+y^{2}=16

STEP 4

Next, we divide each term by16 to get the equation in standard form.
4x216+4y216=1\frac{4x^{2}}{16}+\frac{4y^{2}}{16}=1

STEP 5

implify the equation.
x24+y24=1\frac{x^{2}}{4}+\frac{y^{2}}{4}=1

STEP 6

The center of the ellipse is at the origin (0,0) as there are no xx or yy terms in the equation.

STEP 7

The lengths of the semi-major and semi-minor axes are both 4=2\sqrt{4}=2, so the radius of the ellipse (which is actually a circle in this case) is2.
So, the standard equation of the ellipse is x24+y24=1\frac{x^{2}}{4}+\frac{y^{2}}{4}=1, the center is at the origin (0,0), and the radius is2.

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