Question

Find the equation of the Ellipse given the vertex (0,8) and (0,-8) ; and its eccentricity 3/4

Answer 1

Hi!!

Answer 2

@mathmate

Answer 3

what i know \[\frac{ x^2 }{ a^2 }+\frac{ y^2 }{ b^2 }=1\]

Answer 4

with the eccentricity being 3/4
this means c=3

Answer 5

and b=4 or is it a=4?

Answer 6

but its that the focus points u have there ?

Answer 7

so the vertex is the focus points?

Answer 8

oh, my bad, I didn't read the question well!
yes, so b=8.
and e=3/4=sqrt(1-a^2/b^2) since the major axis is vertical.
So can you solve for a?

Answer 9

Also, e=3/4=c/b, so we can solve for c=3/4*b=6

Answer 10

ok the major axis is the y axis which gives an eccentricity of
c/(major axis) ...c/b
so i taught c would =3 and b=4 so b^2=16
c^=b^2-a^2
9=16-a^2
a^2=-9+16
a^2=7
a=sq.rt 7

Answer 11

The semi-major axis is 8, so b=8.

Answer 12

b=8, c=(3/4)*8=6
a^2=sqrt(b^2-c^2)=...

Answer 13

ok so c=6?

Answer 14

and a=?

Answer 15

a^2=28

Answer 16

a=sq.et 28

Answer 17

Yes, a=sqrt(28), exacly, or sqrt(28)=sqrt(2^2*7)=2sqrt(7).
Both are good.
So now you know both a and b, can you figure out the equation of the ellipse?

Answer 18

so the equation is x^2/28+(y-8)^2/64=1

Answer 19

Yes, that is correct.
We do, however, write it as
x^2/a^2 + y^2/b^2 =1
so take the square-roots and write it as:
x^2/(sqrt(28)^2 + y^2/8^2=1

Answer 20

Sorry, I didn't see the (y-8)^2, there is no translation.
so it should read just y^2.
See the diagram I drew above?
The centre of the ellipse is the mid-point between the foci, namely (0,0).

Answer 21

ok

Answer 22

thanks

Answer 23

you're welcome! :)