Mathematics, 18.06.2020 06:57 lychee1406

Find the center vertices and foci of the ellipse with equation 2x^3+8y^2=16

Answers: 3

Show answers

Answers

Answer from: lyssssasantiago6303

Step-by-step explanation:

First of all, you have a typo. But that's ok, we can work around it. The equation NOT yet in standard form for an ellipse should be:

2x^2+8y^2=16

Putting it into standard form requires that the right side of the equation equals 1. That means that we need to divide everything by 16 to accomplish that. When we do that, we will get the standard form:

$\frac{x^2}{8}+\frac{y^2}{2}=1$

The general form is either

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

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

where a is ALWAYS bigger than b. So if the bigger denominator is under the x numerator, this is an ellipse that is elongated horizontally. If the larger denominator is under the y numerator, then this is an ellipse that is elongated vertically. As you can see from our standard ellipse equation, the 8 is under the x numerator, so this is a horizontally elongated ellipse.

If a^2=8, then $a=\sqrt{8}$ which is approximately 2.83 units.

If b^2=2, then $b=\sqrt{2}$ which is approximately 1.41 units.

Since the ellipse is elongated horizontally, the a value starts at the center of the ellipse and goes 2.83 units to the right and left of the center; the b value starts at the center of the ellipse and goes 1.41 units both up and down from the center. But what's the center?

In the standard form above involving the h and the k in the numerators, the h and k indicate the coordinates of the center. h is like the x coordinate and k is like the y coordinate. Since our ellipse does not have c^2=8-2 a constant in a set of parenthesis for either the x or the y term, the center is at (0, 0). From the center then, we go go both right and left to (2.83, 0) and (-2.83, 0) for the vertices of the ellipse. The co-vertices would be (0, 1.41) and (0, -1.41).

The foci have the formula (h + c, k) for the focal point to the right of the center and (h - c, k) for the focal point to the left of the center. But how do we find c?

c^2=a^2-b^2

We already know that a^2=8 and b^2=2, so

c^2=8-2 and

$c=\sqrt{6}$

That means that the foci have the coordinates $(\sqrt{6},0)$ and $(-\sqrt{6},0)$

and you're done!!

Answer from: Quest

answer: i don't understand. what are the options

step-by-step explanation:

Answer from: Quest

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives -2, 0, 2, 3.

you,

mrb

Other questions on the subject: Mathematics

Mathematics, 21.06.2019 16:30, chloesmolinski0909

Solve the equation w^3 = 1,000 i honestly dont know how to punch this into a calculator pls

Answers: 1

continue

Mathematics, 21.06.2019 17:20, ryleepretty

Two language majors, anna and megan, took exams in two languages. anna scored 85 on both exams. megan scored 74 on the first exam and 85 on the second exam. overall, student scores on the first exam had a mean of 82 and a standard deviation of 4, and the second exam scores had a mean of 71 and a standard deviation of 13. a) to qualify for language honors, a major must maintain at least an 85 average across all language courses taken. so far, which of anna and megan qualify? b) which student's overall performance was better?

Answers: 2

continue

Mathematics, 21.06.2019 20:50, twhalon72

What is the greatest number of parts of a circle that can be formed by cutting the circle with 7 straight cuts? (note: the parts do not have to be equal in size)

Answers: 3

continue

Mathematics, 21.06.2019 22:00, alexisgonzales4752

Find an bif a = {3, 6, 9, 12) and b = {2, 4, 6, 8, 10).

Answers: 1

continue

You know the right answer?

Find the center vertices and foci of the ellipse with equation 2x^3+8y^2=16...