Conics help - eccentricity when a<b (1 Viewer)

[iPhone4]

Hi everyone, for a regular hyperbola with x^2/a^2 -y^2/b^2 = 1,

When a<b, what is the formula to find eccentricity?

Is it the same when a>b?

Usually we use:

b^2=a^2(e^2-1) for a>b.

For a<b, do we use

a^2 = b^2 (e^2-1)?

Likewise for the ellipse, we use:

b^2=a^2 (1-e^2) for a>b

For a<b, do we now use

a^2=b^2 (e^2 -1)?

Thanks (hope this isnt too confusing)

 

[braintic]

Hyperbolas do not work the same as ellipses in this respect.

The orientation of the ellipse depends on which is larger, a or b.

The orientation of the hyperbola depends on whether x^2/a^2 -y^2/b^2 is equal to positive or negative 1. The relative sizes of a and b do not affect the orientation.

So with the ellipse you switch to the alternate eccentricity formula when b>a.
With the hyperbola you switch to the alternate eccentricity formula when the RHS is -1 (or the terms are subtracted the other way around).

So here you use the normal eccentricity formula.

 

[Makematics]

Hyperbolas do not work the same as ellipses in this respect.

The orientation of the ellipse depends on which is larger, a or b.

The orientation of the hyperbola depends on whether x^2/a^2 -y^2/b^2 is equal to positive or negative 1. The relative sizes of a and b do not affect the orientation.

So with the ellipse you switch to the alternate eccentricity formula when b>a.
With the hyperbola you switch to the alternate eccentricity formula when the RHS is -1 (or the terms are subtracted the other way around).

So here you use the normal eccentricity formula.

can they/do they ever ask questions on the alternate hyperbola?

 

[braintic]

can they/do they ever ask questions on the alternate hyperbola?

I don't recall ever seeing one. Not sure whether they can though.

 

[Makematics]

I don't recall ever seeing one. Not sure whether they can though.

they could probably chuck one in, ask you to graph it and claim that it fits in under graphs/transforming known curves or something