SOLUTION: What is the focal point of the ellipse (x+3)^2/16 + (y-1)^2/4 ?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the focal point of the ellipse (x+3)^2/16 + (y-1)^2/4 ?      Log On


   

Question 769001: What is the focal point of the ellipse (x+3)^2/16 + (y-1)^2/4 ?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What is the focal point of the ellipse (x+3)^2/16 + (y-1)^2/4 ?
***
%28x%2B3%29%5E2%2F16+%2B+%28y-1%29%5E2%2F4=1
This is an equation of an ellipse with horizontal major axis.
Its standard form: %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1, a>b, (h,k)=(x,y) coordinates of center
For given ellipse:
center: (-3,1)
a^2=16
b^2=4
c^2=a^2-b^2=16-4=12
c=√12≈3.46 (distance from center to foci on horizontal major axis)
foci: (-3±c, 1)=(-3±3.46,1)=(0.46,1) and (-6.46,1) (note: ellipses have 2 focal points)