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Y^2- 4x² - 4y—8x-16=0

transverse axis ​

1 Answer

4 votes

Answer:

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

Step-by-step 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).

Y^2- 4x² - 4y—8x-16=0 transverse axis ​-example-1
User Robert Prine
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