Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 months ago

# What are the vertices, covertices, and foci of the ellipse (x-5)^2/64 + (y-10)^2/100 = 1?

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(x-5)²/64 + (y-10)²/100 = 1

The denominator of the y term is larger than the denominator of the x term, so the ellipse is vertical.

General equation for a vertical ellipse:

(y-k)²/a² + (x-h)²/b² = 1

with

a ≥ b

center (h,k)

vertices (h,k±a)

co-vertices (h±b,k)

foci (h,k±c), c² = a²-b²

(y-10)²/10² + (x-5)²/8² = 1

center (5,10)

vertices (5,10±10) = (5,20) and (5,0)

co-vertices (5±8,10) = (13,10) and (-3,10)

c = √(10²-8²) = 6

foci (5,10±6) = (5,16) and (5,4) • Login to reply the answers
• Anonymous
7 months ago

a = 5, b = 4

c^2 = 25-16 = 9

c = ±3

Centre = (5,10)

Vertices = (5,0), (5,20)

Co-vertices = (-3,10), (13,10)

Foci (5,13),(5,7)

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• ((x-5)/8)^2 + ((y-10)/10)^2 = 1

Vertices (Major axis): (5,10-10)=(5,0) and (5,10+10)=(0,20)

Co-vertices (Minor axis): (5-8,10)=(-3,10) and (5+8,10)=(13,10)

Foci: (5,10-√(10²-8²))=(5,4) and (5,10+√(10²-8²))=(5,16)

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• (x-h)^2/a^2+ (y-k)^2/b^2 = 1 , with a<b

the vertices (h , k-b) and (h , k+b)

covertices (h-a , k) and (h+a , k)

foci (h , h-c) and (h , h+c) , with c = √(b^2 - a^2)

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