Question

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

Answer 1

Answer:

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

Step-by-step 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)

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

$\tt c = \sqrt{ {a}^{2} - {b}^{2} }$

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

therefore,

foci:(h±c,k)

=(0±6,0)

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

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

$\text{also see the graph}$

Answer 2

Answer:

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

Step-by-step explanation: