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.
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).