**(**2

*x - h*)**+ (**2

*y - k*)**=**2

*r*
where (

*h*,*k*) is the set of coordinates for the center of the circle and*r*is the radius of the circle. We then move on to ellipses, and find the general equation is**[(**2

*x - h*)**] /**2

*a***+ [(**2

*y - k*)**] /**2

*b***= 1**

when

How would one answer the following question?

*a*>*b*(horizontally elongated ellipse; switch the positions of the*x*and*y*terms for a vertically elongated ellipse). The length of the major axis is therefore 2*a*and the minor axis 2*b*. Here is a horizontally-elongated ellipse (image from Wikipedia):How would one answer the following question?

**What are the lengths of the major and minor axes for the following ellipse?**

**4**

*x*2 + 8*x*+ 16*y*2 - 64*y*- 13 = 0

We would first complete the square by adding a total of 68 to both sides (split up as 4 and 64) to reveal two factorable trinomials:

**(4**

*x*2 + 8*x*+ 4) + (16*y*2 - 64*y +*64) - 13 = 68
Factoring out a 4 and 16 from the first and second trinomials respectively, and adding 13 to both sides, this becomes:

**4(**

*x*2 + 2*x +*1) + 16(*y*2 - 4*y +*4) = 81
Factoring the trinomials

**4(**

*x*+ 1)2 + 16(*y*-*2)2 = 81*
and dividing both sides by 81

**(4/81)(**

*x*+ 1)2 + (16/81)(*y*-*2)2 = 1*

This is nearly in the form requisite for an ellipse. How do we rewrite this to show

*a*2*and**b*2 in the denominators? The key is to rewrite the equations first as
[

**(***x*+ 1)2 / (81/4)] + [(*y*-*2)2 / (81/16)] = 1*
(Note: 4/81 = 1 / (81/4) and 16/81 = 1/ (81/16))

**which equals:**[

**(**

*x*+ 1)2 / (9/2)**2**

**] + [(**

*y*-*2)2 / (9/4)***2**

**] = 1**

This ellipse, centered at (-1, 2), has major axis length (2

*a*) of**2(9/2) =****9**whereas the minor axis (2*b*) measures**2(9/4) =**9/2.

For an informative website on conic sections, see http://en.wikipedia.org/wiki/Conic_section

For online tutoring help, go to Virtual Tutor Center at http://www.virtualtutorcenter.com

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