How do I find the center, vertices, foci, and eccentricity of the ellipse? x2+8y2−8x−16y−40=0

1 Answer
Douglas K.
Dec 2, 2017

One should complete the squares so that the equation may be written in one of the two following forms:

(x−h)2a2+(y−k)2b2=1 [1]

(y−k)2a2+(x−h)2b2=1 [2]

where a>b

Explanation:

Given:

x2+8y2−8x−16y−40=0

Add 40 to both sides:

x2+8y2−8x−16y=40

Group the x terms and the y terms together:

(x2−8x)+(8y2−16y)=40

We cannot complete the square unless the leading coefficient is 1, therefore, we remover a factor of 8 from the y terms:

(x2−8x)+8(y2−2y)=40

Because (x−h)2=x2−2hx+h2 we want to insert an h2 into the x term's group but we must, also add h2 to the right side so that equality is maintained:

(x2−8x+h2)+8(y2−2y)=40+h2

Matching the x terms with the general pattern, (x−h)2=x2−2hx+h2, we observe that the equation

−2hx=−8x

will allow us to solve for the value of h:

h=4

This means that h2 on the right side becomes 16 and the group of x terms become (x−4)2

(x−4)2+8(y2−2y)=40+16

Combine like terms:

(x−4)2+8(y2−2y)=56

We want insert k2 into the y terms but to maintain equality we must add 8k2 to the right side:

(x−4)2+8(y2−2y+k2)=56+8k2

Matching the y terms with the general pattern, (y−k)2=y2−2ky+k2, we observe that the equation

−2ky=−2y

will allow us to solve for the value of k:

k=1

This means that 8k2 on the right side becomes 8 and the group of y terms become (y−1)2

(x−4)2+8(y−1)2=56+8

Combine like terms:

(x−4)2+8(y−1)2=64

Divide both sides by 64:

(x−4)264+(y−1)28=1

Write the denominators as squares:

(x−4)282+(y−1)2(2√2)2=1

This is the same form as equation [1],

(x−h)2a2+(y−k)2b2=1 [1]

The center, (h,k)=(4,1)
The vertices are, (h−a,k)=(−4,1) and (h+a,k)=(12,1)
The foci are, (h−√a2−b2,k)=(4−√56,1) and (h+√a2−b2,k)=(4+√56,1)
The eccentricity is √a2−b2a=√568

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