Hyperbolas equivalent of an ellipsis's minor axis?

This is algebra. I think I missed this part in class. I know a Hyperbola has a conjugate axis, but is this the hyperbolas equivlent of an elipise's minor axis? If so, how do you find the measurement? I'm writing a forumla for a hyprbola in standard form and need to find b^2. Thanks.

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HallsofIvy

Homework Helper

In particular, if you have an ellipse given by x^2/a^2+ y^2/b^2= 1,
the ellipse is contained in a rectangle with length and width 2a and 2b. If you extend the diagonals of that rectangle, you get the asymptotes of a hyperbola with equation x^2/a^2- y^2/b^2= 1 (as well as y^2/b^2- x^2/a^2= 1).

It's not clear from what you write what information you are given to identify the hyperbola (and thus find its equation).

Assuming that the hyperbola has center at (0,0) and axes of symmetry along the x and y axes, then it can be written x^2/a^2- y^2/b^2= 1 (0r y^2/b^2- x^2/a^2= 1).

I will assume that you know "a"- that is, that you know where the hyperbola crosses the x-axis: (a,0) and (-a,0) and that you know the asymptotes (There are an infinite number of different hyperbolas that go through (a,0),(-a,0) with different asymptotes).

To find "b", draw vertical lines through (-a,0) and (a,0):that is, draw the lines x=a and x=-a. Where they cross the asymptotes, draw horizontal lines:y= "something". That rectangle is the rectangle I referred to earlier and that "something" is b.

Anyway, this hyperbola is of the form y=k/x, so it's asymtotes are the x and y axis. You know this because it's foci were given as (8,8) and (-8,-8). My teacher gave the focal constant as 8, which won't work, she made a mistake. Naturally, it should be 16. So, lets assume that she gave the focal constant to be 16 because I want to pratice this. This information is all she gave me. When you have a hyperbola of the equation y=k/x, the vertices, I believe to be (square root of K,square root of K) and (-square root of K,-square root of K). I just found this out today in class, so I've answered that part of my question. But I still don't understand how to find the length of the hyperbolas congugate axis when it's in the form y=k/x.

When you have a hyperbola of the form (x^2/a^2)-(y^2/b^2)=1, the hyperbola probably (I'm saying probably cause my teacher wouldn't give a hyperbola with is asymptotes as the x and y axis in this form) doesn't have the x and y axis as it's asymptoes, and isn't reflection symetric to the line y=x, so all those forumlas don't apply. So, just knowing the where the foci are and what the focal constant is, can one find the vertices, and the lengths of the 2 axis?

Thanks!

HallsofIvy

Homework Helper
A hyperbola satisfying y= k/x (and since you are given that it passes through (8,8) and (-8,-8), clearly k must be 64) has y= x as an axis. I was going to say it's vertices are at the intersection of y= x with y= k/x but obviously those are the (8,8) and (-8,-8) points you are given.

Remember I suggested drawing a rectangle with sides passing through the vertices and the asymptotes as diagonals? In this case, two sides of that rectangle would have to pass through (8,8) and (-8,-8) and be perpendicular to y= x which means parallel to y= -x: they have to have slope -1. The line with slope -1 passing through (8,8) is y= -(x-8)+8= -x+16 and the line with slope -1 passing through (-8,-8) is y= -(x+8)- 8= -x-16. y= -x+16 crosses the asymptote y= 0 (x-axis) when x= 16 and the asymptote x=0 (y-axis) when y= 16. Of course, the other line crosses the axes at x= -16 and y= 16.

The "rectangle" I consider so important has vertices at (16,0), (0,16), (-16,0), and (0,-16). The length of those sides is, of course, sqrt(16^2+ 16^2)= 16 sqrt(2). The "a" and "b" you want are half that: 8 sqrt(2). Notice that the "rectangle" is actually a square. You could just have calculated that the distance from (0,0), then center of the hyperbola, to (8,8), one of the vertices, is 8 sqrt(2). If you were to rotate the axes 45 degrees so that there were along the axes of the hyperbola, the equation would be change to x^2/- y^2/32= 1. The focal length (distance from the center to the focus) is, by the way, sqrt(a^2+ b^2)= sqrt(128)= 16 as you said.

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