Need help in determining if this is possible (Conic Section)

[AhrenMath]

The problems below were given by my teacher however, I noticed that the y in the foci and vertices aren't the same thus, making me think that either one of them is a typo or error. I'm wondering whether if the given is faulty or possible to solve as a hyperbola. Also, #2 has the same problem with different y.(Note: I tried solving them but treated 13 as 3 and -1 as 1)

1)Hyperbola with foci at (2 +- √13,13), vertices at (5,3) and (-1,3)

2)Hyperbola with C(-2,1) , focus at (6, -1) and transverse axis of length 8.

 

[Deleted member 4993]

The problems below were given by my teacher however, I noticed that the y in the foci and vertices aren't the same thus, making me think that either one of them is a typo or error. I'm wondering whether if the given is faulty or possible to solve as a hyperbola. Also, #2 has the same problem with different y.(Note: I tried solving them but treated 13 as 3 and -1 as 1)

1)Hyperbola with foci at (2 +- √13,13), vertices at (5,3) and (-1,3)

2)Hyperbola with C(-2,1) , focus at (6, -1) and transverse axis of length 8.

I have not worked out the solutions - but problems "seem" to be correct (solvable).

 

[AhrenMath]

I have not worked out the solutions - but problems "seem" to be correct (solvable).

Well to be honest I'm actually surprised. If I may ask does this use a different formula aside from the usual (x-h)^2/a^2 - (y-k)^2/b^2 = 1? If so how do you get the vertices and the others like foci?

 

[mmm4444bot]

… does this use a different formula aside from the usual

(x-h)^2/a^2 - (y-k)^2/b^2 = 1 …

That's the formula for a hyperbola that opens left and right.

The formula for a hyperbola that opens up and down is similar:

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

I have not looked at the second exercise, but I considered that maybe the axes were rotated on the first one. But, having plotted the given points, I don't see how a hyperbola can be drawn through those vertices (blue) with such foci (red). :?

 

[mmm4444bot]

Note: I tried solving them but treated 13 as 3 and -1 as 1

Had time for more. I think we can define a hyperbola with the givens in the second exercise, but the transverse axis would have negative slope (i.e., a rotated hyperbola).

PS: On the first one, you could also treat 3 as 13. On the second (assuming you're not rotating hyperbolas) you could also treat 1 as -1. :cool:

 

[AhrenMath]

Had time for more. I think we can define a hyperbola with the givens in the second exercise, but the transverse axis would have negative slope (i.e., a rotated hyperbola).

PS: On the first one, you could also treat 3 as 13. On the second (assuming you're not rotating hyperbolas) you could also treat 1 as -1. :cool:

Thanks. I really do think this is an error on the teachers part. I appreciate the help a lot as me and my other classmates have been baffled by this equation for the past few days.