Taylor B.

asked • 04/09/14# Writing Hyperbolas in Standard Form? 2 Questions

I am given the vertices (14,-6),(-8,-6) and the endpoints of the conjugate axis, (3,1) and (3,-13)

I also need to write the standard form of a hyperbola given the center (-2,-4) and it says that the transverse axis is vertical and 14 units long. The conjugate axis is 24 units long.

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## 1 Expert Answer

Steve S. answered • 04/09/14

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Tutoring in Precalculus, Trig, and Differential Calculus

The transverse axis is the segment connecting the vertices, so it’s horizontal and (14-(-8)) = 22 units long.

The conjugate axis is vertical and (1-(-13)) = 14 units long.

The center is the midpoint of the transverse axis:

C((14-8)/2,(-6-6)/2) = C(3,-6)

The proper equation format is:

((x-h)/a)^2 – ((y-k)/b)^2 = 1

(h,k) is the center. a and b are half the lengths of transverse and conjugate axes, respectively.

So the proper equation is:

((x-3)/11)^2 – ((y+6)/7)^2 = 1

2.) I also need to write the standard form of a hyperbola given the center (-2,-4), and it says that the transverse axis is vertical and 14 units long. The conjugate axis is 24 units long.

The proper equation format is:

((y-k)/b)^2 - ((x-h)/a)^2 = 1

(h,k) is the center. a and b are half the lengths of conjugate and transverse axes, respectively.

So the proper equation is:

((y+4)/7)^2 - ((x+2)/12)^2 = 1

See GeoGebra sketches here: http://www.wyzant.com/resources/files/268760/hyperbolas

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Taylor B.

04/10/14