Answer:
[tex]\boxed{\boxed{C) Center:(0,0);\: Vertices:(-10,0),(10,0);\:Foci:(-6,0),(0,6)}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
- conic geometry
- algebra
- PEMDAS
given:
- [tex] \frac{ {x}^{2} }{100} + \frac{ {y}^{2} }{64} = 1[/tex]
to find:
tips and formulas:
- [tex] \sf ellipse \:equation : \\ \tt\frac{( {x}^{2} - h)}{ {a}^{2} } + \frac{( {y}^{2} - k)}{ {b}^{2} } = 1[/tex]
- [tex] \sf c(h,k)[/tex]
- [tex] \sf v(h \pm a,k)[/tex]
- [tex]\sf F(h\pm c,k)[/tex]
- [tex] \sf c = \sqrt{ {a}^{2} - {b}^{2} } [/tex]
let's solve:
- [tex] \sf rewrite \:the \: given \: equation \: as \: ellipse \: equation \\ \tt \frac{ {(x - 0)}^{2} }{100} + \frac{ ({y - 0)}^{2} }{64} = 1[/tex]
since we can see that h and k is 0
therefore
c(0,0)
[tex] \sf vertices :( h \pm a,k)[/tex]
given:a²=100
let's find a
a=√100
a=10
therefore
vertices:(0±10,0)
=(0+10,0) and (0-10,0)
=(10,0) and (-10,0)
[tex] \sf foci : (h \pm c,k)[/tex]
[tex] \tt c = \sqrt{ {a}^{2} - {b}^{2} } [/tex]
[tex] \sf \: c = \sqrt{ {10}^{2} - {8}^{2} } \\ \sf \: c = \sqrt{(10 + 8)(10 - 8)} \\ \sf c = \sqrt{(18)(2)} \\ \sf c = \sqrt{36} \\ \tt c = 6 [/tex]
therefore,
foci:(h±c,k)
=(0±6,0)
=(0+6,0) and (0-6,0)
=(6,0) and (-6,0)
[tex]\text{also see the graph} [/tex]
