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