How do you find the center, vertices, foci and eccentricity of #4x² + 9y² - 8x - 36y + 4 = 0#?

1 Answer
Roella W. · Stefan V.
Feb 5, 2016

The centre is at #(1,2)#
The vertices are at #(1/2,2 ), (3/2,2), (1,2*1/3) # and #(1,1*2/3)#
The foci are at #(1-sqrt(5)/6, 2)# and #(1+sqrt(5)/6,2)#

The eccentricity is # (sqrt(5)/6)#

Explanation:

Rearrange the elements to get the standard form of an ellipse
#(x-h)^2/a^2 + (y-k)^2/b^2 = 1# where #(h,k)# is the centre and #a# and #b# are the semi-minor and semi-major axes respectively.

So in this example
#4x^2 -8x +9y^2 - 36y +4 = 0#

Using the method of completing squares

#4(x^2 - 2x) + 9 (y^2-4y) +4 = 0#
#4(x-1)^2 -1 +9(y-2)^2 - 4 + 4 = 0#
#4(x-1)^2 +9(y-2)^2 = 1#
#(x-1)^2/(1/4) +(y-2)^2/(1/9)=1#

Therefore the centre is at #(1,2)#
The semi-major axis has length #1/2# and the semi-minor axis has length #1/3# so the vertices are at #(1/2,2 ), (3/2,2), (1,2*1/3) # and #(1,1*2/3)#

If #c# is the distance between the foci,
#1/4-c^2=1/9#
#:.c^2=5/36# and #c = sqrt(5)/6#

The foci are at #(1-sqrt(5)/6, 2)# and #(1+sqrt(5)/6,2)#

The eccentricity is #c/a = (sqrt(5)/6)#

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