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Elyse Keith
Chemistry
21 July, 17:38
Y^2 - 4x² - 4y-8x-16=0
transverse axis
+1
Answers (
1
)
Joel Padilla
21 July, 18:02
0
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).
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