Question 628918: A focal point of the elipse  ((x+3)^2)/16)+(y-1)^2)/4)=1 is:
a.) (-1,1)
b.) (-3-2 radical 3, 1)
c.) (-7,1)
d.) 3+2 radical 3, 1)
* note: when ever i have entered the equation in to graph it it keeps saying 1/16 (x+3) but that is not correct it is x+3^2 all over 16.
I have worked it some way but I dont know how to continue it. I have:
Centre: (-3,1)
a=16
b=3
c=15.716
but i dont know how to find the focal point.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find a focal point of the ellipse:
((x+3)^2)/16)+(y-1)^2)/4)=1
This is an equation of an ellipse with horizontal major axis:
Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), (h,k)=(x,y) coordinates of center
For given ellipse:
center(-3,1)
a^2=16
a=√16=4
b^2=4
b=√4=2
c^2=a^2-b^2=16-12=4
c=√4=2
Foci: (Both focal points)=(-3±c,1)=(-3±2,1)=(-5,1) and (-1,1)
answer a is the correct answer
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