# HELPP! minor axis of ellipse

• Dec 10th 2008, 05:21 PM
emetty90
HELPP! minor axis of ellipse
How would i go about finding the minor axis of the ellipse:
4x^2+y^2-32x+6y+57=0
Thanks
• Dec 10th 2008, 05:38 PM
skeeter
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.
• Dec 10th 2008, 05:53 PM
emetty90
Why did you add 64 to the right side?
• Dec 10th 2008, 06:03 PM
skeeter
Quote:

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.