How do you find the coordinates of the center, foci, the length of the major and minor axis given #16x^2+25y^2+32x-150y=159#?

1 Answer
Narad T.
Nov 2, 2016

The center is #=(-1,3)#
The foci are #(2,3)# and #(-4,3)#
The length of the major axis is #=10#
and the length of the minor axis is #=8#

Explanation:

Rewriting the equation
#16(x^2+2x)+25(y^2-6y)=159#
#16(x^2+2x+1)+25(y^2-6y+9)=159+16+225#
#16(x+1)^2+25(y-3)^2=400#
#(x+1)^2/25+(y-3)^2/16=1#
Comparing this to the stadard form of the ellipse
#(x-x_0)^2/a^2+(y-y_0)^2/b^2=1#
The center is #(x_0,y_0)=(-1,3)#
To calculate the foci, we need #c=sqrt(a^2-b^2)#
#c=sqrt(25-16)=3#
And the foci are #(x_0+-c,y_0)#
This lead us to F#(2,3)# and F'#(-4,3)#
The length of the major axis is #=10#
and the length of the minor axis is #=8#

graph{(x+1)^2/25+(y-3)^2/16=1 [-11.625, 10.875, -2.865, 8.385]}

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