HELPP! minor axis of ellipse

**emetty90** · December 10th 2008, 04:21 PM

How would i go about finding the minor axis of the ellipse:
4x^2+y^2-32x+6y+57=0
Thanks

**skeeter** · December 10th 2008, 04:38 PM

complete the square for both x and y ...

$4x^2+y^2-32x+6y+57=0$

$4x^2 - 32x + y^2 + 6y = -57$

$4(x^2 - 8x) + y^2 + 6y = -57$

$4(x^2 - 8x + 16) + y^2 + 6y + 9 = -57 + 64 + 9$

$4(x - 4)^2 + (y + 3)^2 = 16$

$\frac{(x-4)^2}{4} + \frac{(y+3)^2}{16} = 1$

$\frac{(x-4)^2}{2^2} + \frac{(y+3)^2}{4^2} = 1$

the last equation contains information that the length of the semi-minor axis is 2.

**emetty90** · December 10th 2008, 04:53 PM

Why did you add 64 to the right side?

**skeeter** · December 10th 2008, 05:03 PM

Originally Posted by emetty90

Why did you add 64 to the right side?

because in the expression $4(x^2 - 8x + 16)$ on the left side of the equation, the 4 distributes to the added 16, resulting in an extra 64 on the left side, which must be balanced on the right side to maintain the equality.

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