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shtirl [24]
3 years ago
13

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

Mathematics
2 answers:
anzhelika [568]3 years ago
7 0

Answer:

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

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • conic geometry
  • algebra
  • PEMDAS
<h3>given:</h3>
  • \frac{ {x}^{2} }{100}  +  \frac{ {y}^{2} }{64}  = 1
<h3>to find:</h3>
  • center
  • vertices
  • foci
<h3>tips and formulas:</h3>
  • \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} }
<h3>let's solve:</h3>
  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 =  \sqrt{(10  +  8)(10 - 8)}  \\  \sf c =  \sqrt{(18)(2)}  \\  \sf c =  \sqrt{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}

elena55 [62]3 years ago
3 0

Answer:

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

Step-by-step explanation:

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