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Question 826098: determine the equation of the hyperbola whose vertices are the foci of the ellipse 11x^2 + 7y^2 + 14y - 70 = 0 and its foci are the vertices of the given ellipse.
thanks :D
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! determine the equation of the hyperbola whose vertices are the foci of the ellipse 11x^2 + 7y^2 + 14y - 70 = 0 and its foci are the vertices of the given ellipse.
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11x^2 + 7y^2 + 14y - 70 = 0
complete the square:
11x^2+7(y^2+2y+1)=70+7=77
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 of th ellipse:
center:(0,-1)
a^2=11
a=√11≈3.32
vertices: (0,-1±a)=(0,-1±√11)=(0,-1±3.32)=(0,-4.32)and(=(0,+2.32)
b^2=7
b=√7
c^2=a^2-b^2=11-7=4
c=±2
foci:(0,-1±c)=(0,-1±2)=(0,-3)and(=(0,+1)
..
For given hyperbola:
assume same center as given ellipse: (0,-1)
Hyperbola has a vertical transverse axis:
Its standard form of equation: 
a=2(foci of ellipse)
a^2=4
c^2=11(vertex of ellipse)
c^2=a^2+b^2
b^2=c^2-a^2=11-4=7
Equation of given hyperbola:
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