Question

I'm having a really hard time with this one... Does anyone know how to do this?

Find the vertices, foci, and eccentricity of the ellipse.

x^2 = 36 − 6y^2

Answer 1

the e value is c/a as i recall from yesterday :)

Answer 2

x= + or - 6 y=0

Answer 3

So is the vertices (0,0) ?

Answer 4

-6,0 or 6,0

Answer 5

turn this into the normal looking ellipse equation

x^2 = 36 − 6y^2 ; +6y^2
+6y^2 +6y^2
----------------

x^2 + 6y^2 = 36 ; divide out the 36
/36 / 36 /36
----------------

x^2 y^2
--- + --- = 1
36 6

Answer 6

rate gud answer plzz

Answer 7

What is the eccentricity?

Answer 8

In mathematics, the eccentricity, denoted e or , is a parameter associated with every conic section. It can be thought of as a measure of how much the conic section deviates from being circular.
In particular,
The eccentricity of a circle is zero.
The eccentricity of an ellipse which is not a circle is greater than zero but less than 1.
The eccentricity of a parabola is 1.
The eccentricity of a hyperbola is greater than 1.

Answer 9

to determine eccentricity we have to know what this form tells us; the larger value is the major axis and is by convention called: a^2

the smaller is called b^2

Answer 10

Any conic section can be defined as the locus of points whose distances are in a constant ratio to a point (the focus) and a line (the directrix). That ratio is called eccentricity, commonly denoted as "e."
The eccentricity can also be defined in terms of the intersection of a plane and a double-napped cone associated with the conic section. If the cone is oriented with its axis being vertical, the eccentricity is

where α is the angle between the plane and the horizontal and β is the angle between the cone and the horizontal.
The linear eccentricity of a conic section, denoted c (or sometimes f or e), is the distance between its center and either of its two foci. The eccentricity can be defined as the ratio of the linear eccentricity to the semimajor axis a: that is, .

Answer 11

a is the hypotenuse of a rt triangle formed from the vertex lengths and the foci

Answer 12

What on earth? I'm even more confused now lol

Answer 13

rate gud answer plzz

Answer 14

I have done problems like this in the past I am just having a really hard time with this one for some reason... It's not that I don't know how to do them I just am not coming up with a good answer for this one. The answers I am getting to not seem right

Answer 15

a^2 = c^2 + b^2 is how we determine c :)

sqrt(36) = c^2 + sqrt(6)

6-sqrt(6) = c^2

c = \(\sqrt{6-\sqrt{6}}\) if i did it right

Answer 16

the e value is defined by our c/a so we just plug those in

Answer 17

Wouldn't C just be -6 and 6 without the square root

Answer 18

you may be right :)

Answer 19

im old and get confused so it helps when you correct my mistakes :)

Answer 20

36 = c^2 + 6

30 = c^2

sqrt(30) = c looks much niver ;)

Answer 21

It's fine!

So A=0
B= -6, 6
C= -6, 6

?

Answer 22

Would that make the eccentricity 0?

Answer 23

a b and c are just conventional names for parts of the ellipse; they are single numbers

Answer 24

a is the length of the major axis vertex from the center; b is the length of the minor axis from the center; c is the distance from center to foci

Answer 25

c and b make the legs of a right triangle and a is the hypotenuse

Answer 26

I am working with ellipses though

Answer 27

I'll figure it out... thanks

Answer 28

ellipse
foci | ((-sqrt(30), 0) | (sqrt(30), 0))~~((-5.47723, 0) | (5.47723, 0))
center | (0, 0)
semimajor axis length | 6
semiminor axis length | sqrt(6)~~2.44949
area | 6 sqrt(6) pi~~46.1718
perimeter | 24 E(5/6)~~27.7302
focal parameter | sqrt(6/5)~~1.09545
eccentricity | sqrt(5/6)~~0.912871

Answer 29

x^2 y^2
--- + --- = 1 ; is how an ellipse is set up; a and b can change position
a^2 b^2 tho depending on major minor axis

Answer 30

a is major and b is minor

a^2 = b^2 + c^2 is the pythag thrm for the relationship between them

36 = 6 + c^2

30 = c^2

sqrt(30) = c
....................................

e value is defined as c/a

sqrt(30)
------- = sqrt(5/6) if i see it right
sqrt(36)

Answer 31

thanks!