Y^2 - 4x² - 4y-8x-16=0

transverse axis

line segment of length 8 between the vertices at (-1, - 2) and (-1, 6)

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).