# Thread: HELPP! minor axis of ellipse

1. ## 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

2. complete the square for both x and y ...

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

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

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

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

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

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

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

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

3. Why did you add 64 to the right side?

4. Originally Posted by emetty90
Why did you add 64 to the right side?
because in the expression $\displaystyle 4(x^2 - 8x + 16)$ 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.