Question 542265: What are the vertices and the foci of this conic.
9x^2+ 4y^2 + 64y = -220
I basically need the most help with how to solve it, not just simply the answer please.
Thank you
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What are the vertices and the foci of this conic.
9x^2+ 4y^2 + 64y = -220
I basically need the most help with how to solve it, not just simply the answer please.
**
9x^2+ 4y^2 + 64y = -220
complete the square
9x^2+4(y^2+16y+64)=-220+256=36
9x^2+4(y+8)^2=36
divide by 36
x^2/4+(y+8)^2/9=1
This is an equation of an ellipse with vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1,a>b, with (h,k) being the (x,y) coordinates of the center.
..
For given problem:
Center:(0,-8)
a^2=9
a=√9=3
vertices: (0,-8±a)=(0,-8±3)=(0,-11) and (0,-5)
..
b^2=4
..
c^2=a^2-b^2=9-4=5
c=√5≈2.24
Foci: (0,-8±c)=(0,-8±√5)=(0,-10.24) and (0,-5.76)
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