Question

Please help!!
Find the center, vertices, and foci of the ellipse with equation

Answer 1

center (0;0), vertices (0;10) and (0;-10), foci (0;6) and (0;-6)

Explanation:

Answer 2

`\boxed{\boxed{C) Center:(0,0);\: Vertices:(-10,0),(10,0);\:Foci:(-6,0),(0,6)}}`

Explanation:

to understand this

you need to know about:

• conic geometry
• algebra
• PEMDAS

given:

• 
  `\frac{ {x}^(2) }{100} + \frac{ {y}^(2) }{64} = 1`

to find:

• center
• vertices
• foci

tips and formulas:

• 
  `\sf ellipse \:equation : \\ \tt\frac{( {x}^(2) - h)}{ {a}^(2) } + \frac{( {y}^(2) - k)}{ {b}^(2) } = 1`
• 
  `\sf c(h,k)`
• 
  `\sf v(h \pm a,k)`
• 
  `\sf F(h\pm c,k)`
• 
  `\sf c = \sqrt{ {a}^(2) - {b}^(2) }`

let's solve:

1. 
   `\sf rewrite \:the \: given \: equation \: as \: ellipse \: equation \\ \tt \frac{ {(x - 0)}^(2) }{100} + \frac{ ({y - 0)}^(2) }{64} = 1`

since we can see that h and k is 0

therefore

c(0,0)

`\sf vertices :( h \pm a,k)`

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)

`\sf foci : (h \pm c,k)`

`\tt c = \sqrt{ {a}^(2) - {b}^(2) }`

`\sf \: c = \sqrt{ {10}^(2) - {8}^(2) } \\ \sf \: c = √((10 + 8)(10 - 8)) \\ \sf c = √((18)(2)) \\ \sf c = √(36) \\ \tt c = 6`

therefore,

foci:(h±c,k)

=(0±6,0)

=(0+6,0) and (0-6,0)

=(6,0) and (-6,0)

`\text{also see the graph}`