Question

Find the vertices, foci, and eccentricity of the ellipse.
5x^2+y^2=6
can someone help me with this?

Answer 1

when x = 0, y = +-sqrt(6)

when y = 0, x = +-sqrt(6/5)

the rest of it is greek to me at the moment :)

Answer 2

it appears to be taller than wider

Answer 3

those might be the vertices :)

Answer 4

i get how to do the problem. It should look like x^2/6/5 + y^2/30 = 1. I just dont know how to clear the fraction under the x^2 term.

Answer 5

1/(6/5) just means find the reciprocal of 6/5

Answer 6

the equation must stay equal to 1.

Answer 7

foci are c^2 = a^2 - b^2....

Answer 8

[(x-h)^2]/a+[(y-k)^2}/b=1 what is this form?

Answer 9

5^2 - 1^2 = 24.... is the slant = sqrt(24)? cant recall to much about it

Answer 10

i remember writing ellipses in that form

Answer 11

do the foci need to be on the x axis?

Answer 12

yes because it is a horizontal equation

Answer 13

heres some information on horizontal ellipses is x^2/a^2 + y^2/b^2 = 1
a> b > 0
vertices (+ or -a,0)
foci (+ or - c,0)

Answer 14

I know everything just not how to clear that 6/5 without changing the = 1 part.

Answer 15

if i could just put 5x^2+y^2=6 in the x^2/a^2 + y^2/b^2 = 1 form. i could solve it.

Answer 16

I would just leave the 6/5 where it is. But I think there is a 6 under the y^2, not 30.

Answer 17

your right

Answer 18

What do you mean "solve"? You say you know how to do it...

Answer 19

I mean find all the information from it and then graph it.

Answer 20

Okay, the vertices are (+-sqrt(6/5), 0) and (0, +-sqrt(6)). The foci you find by taking c^2 = a^2 - b^2. So, c^2 = 6 - 6/5 = 24/5. c = +-sqrt(24/5). The foci are on the y-axis because the ellipse is vertically oriented: (0, +-sqrt(24/5)). The eccentricity you find by taking c / a, so sqrt(24/5) / sqrt(6) = sqrt(4/5).

Answer 21

Cool. thats what i got. Thanx!