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September 20, 2014

September 20, 2014

Posted by **Sheila** on Monday, March 28, 2011 at 3:08pm.

(x-3(^2/49 + (y-4)^2/4 = 1

Center would be (3,4)

A=7

b=2

Vertices would be found this way:

(h+-a,k)

(3-7,4)

(3+7,4)

Vertices = (-4,4)

(10,4)

Co-vertices

(h,k+-b)

(3,4-2)

(3,4+2)

Co-vertices=(3,2)

(3,6)

If ound "c" by 49-4 = c^2

45=c^2

3sqrt(5)=c

6.7=c

Foci =(h+-c,k)

(3-6.7,4)

(3+6.7,4)

Co-vertices=(-3.7,4)

(9.7,4)or left in radical it would be: 3- (3sqrt5),4

3+(3sqrt5),4

- Algebra II-Please check my calculation -
**Reiny**, Monday, March 28, 2011 at 4:33pmI agree with all your answers.

- Algebra II-Please check my calculation -
**Sheila**, Monday, March 28, 2011 at 5:55pmThank you

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