Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±2 divided by 3.x.

Answer 1

standard form for a hyperbola is -x²/a² + y²/b² = 1 for when the hyperbola has a major axis on the y axis, . with asymptotes = y=bx/a. when x=0, and y=4, 4²/b² = 1 and b= 4, then y=4x/6=2x/3 a=6, b=4 substitute into the standard form

-x²/36 + y²/16 = 1

Answer 2

Hello Joslen,

The vertices (0,±4) are on the y-axis, and the center of the hyperbola will be midpoint of the line segment joining the two vertices. Thus, the center is (0,0). The standard form of the equation of a hyperbola with center at the origin (0,0) and vertices (0,±b) is given by

y²/b² - x²/a² = 1

where a and b are positive constants. Since the vertices in your problem are (0,±4), we have b = 4. Also, for the equation in the form above, it can be shown that the asymptotes of the hyperbola are given by

y = ±(b/a)x

Since the asymptotes in your problem are the lines y = ±(2/3)x, we can write

b/a = 2/3.

Now substitute b = 4 and solve the above for a.

4/a = 2/3

2a = 12

a = 6

Therefore, the equation of the hyperbola is

y²/4² - x²/6² = 1, or

y²/16 - x²/36 = 1

Hope that helps! Let me know if you need any further explanation.

William