Y^2- 4x² - 4y—8x-16=0
transverse axis

Explanation:

The equation can be rearranged to standard form.

  (y^2 -4y) -4(x^2 +2x) = 16

  (y^2 -4y +4) -4(x^2 +2x +1) = 16 +4 -4

  (y -2)^2 -4(x +1)^2 = 16

  (y -2)^2 /16 -(x +1)^2/4 = 1

This is of the form ...

  (y -k)^2/a^2 -(x -h)^2/b^2 = 1

where the transverse axis is 2a and the center is (h, k). Here, a=4, so 2a = 8.

The transverse axis is a vertical line segment of length 8, centered on (-1, 2).