Ellipse(Finding the center, vertices)

Question

So this is the equation 16x^2 + 9y^2 = 144

So this is what I did:

16x^2/144 + 9y^2/144 = 144/144

x^2/9 + y^2/16 = 1

a^2= 9 ; a = 3

b^2= 16 ; b=4

so if I solve for the c

c^2 = a^2 - b^2

c^2 = 9 - 16

c^2 = -7 ; c= √-7

So from that, what becomes my foci and vertices?

Or did I do wrong from my solving.

King Squirrel · Accepted Answer · 2014-05-08T16:50:40.290

0

a^2 is not 9, it is 16. A is defined to be the larger number. So a=4 and b=3. Therefore:

c^2=16-9

c^2=7

c=+-sqrt(7)

For the vertices:

The vertices are defined to be the points

(0,+-a) or (+-a,0)

In this case when x=0, y is +-4 so the vertices are (0,+-4).

edited May 08 '14 at 16:50

answered May 08 '14 at 16:41

King Squirrel

So if that's the case I will have an equation of y^2/a^2 + x^2/b^2 = 1. And the foci will be on the y axis? Right? – Z'K May 08 '14 at 16:52
Exactly right, as well as the vertices. – King Squirrel May 08 '14 at 17:05
1

Thank you so much. That really helps a lot to me :) – Z'K May 08 '14 at 17:06
The King orders Z'K to choose the King's answer by clicking the check mark and up arrow. – King Squirrel May 08 '14 at 17:08
I can't up the arrow :'( it still needs 15 rep. I only got 10 rep – Z'K May 08 '14 at 17:09
The King pardons Z'K. – King Squirrel May 08 '14 at 17:09

1 Answers1