Question 689299: how do you find the vertices, foci, of the ellipse.
56x^2 + 7y^2 = 8
56x^2/8+7y^2/8=8/8
x^2/sqrt7+y^2/(sqrt14/4)^2=1
eccentricity=sqrt14/4
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! how do you find the vertices, foci, of the ellipse.
56x^2 + 7y^2 = 8
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56x^2+7y^2 = 8
7x^2+7y^2/8 = 1
This is an equation of an ellipse with vertical major axis.
Its standard form: , a>b, (h,k)=(x,y) coordinates of center
For given equation:7x^2+7y^2/8 = 1
center: (0,0)
a^2=8/7
a=√(8/7)≈1.07
vertices:(0,0±a)=(0,0±1.07)=(0,-1.07) and (0,1.07 )
b^2=1/7
b=√7/7≈.38
c^2=a^2-b^2=8/7-1/7=7/7=1
c=1
foci: (0,0±c)=(0,0±1)=(0,-1) and (0,1)
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