Khan/khan-exercises — Issue #5188: Equation of a hyperbola - ± ambiguity
Author: KhanBugz
Created: 2011-10-29 21:25:38+00:00
State: open
Labels: safari, snowleo

Reporter: rhaodess@yahoo.com

I know this may seem nit picky, but just to let y'all know. The plus or minus sign (±) should not operate on the vertical shift term. For example

[(x-5)^2]/4 - [(y-3)^2]/81 = 1 :gives asymptotes

y = [±9(x-5)/2] + 3 ; notice parentheses isolate the ± sign on the 9(x-5)/2 term only. The +3 term must remain +3, such that:

y= [9(x-5)/2] + 3
or
y= [-9(x-5)/2] + 3

If we have
y = ±[9(x-5)/2 + 3]
which is what it looks to be in the practice exercise we get
y= [9(x-5)/2 + 3]
or
y= [-9(x-5)/2 - 3]
This would be wrong.

I understand that if I write it as:
y=±9(x-5)/2 + 3; it might make sense but by the commutative law this can also be written as:
y=±[3+9(x-5)/2] and this is accepted as correct in the exercise.
y=±3+9(x-5)/2; implies

y = 3+9(x-5)/2
or
y = -3+9(x-5)/2
again wrong.
You might consider creating a separate input box for the vertical shift part. Something like the following, such that the

y = [±value] + [vertical shift value].

If it shifts down, the student would need to put a negative sign on it, if it shifts up, just leave it positive. so for example problem I used, it would look like:
y = [±9(x-5)/2] + [3]; student inserts 3.

If we had the equation

[(x-5)^2]/4 - [(y+3)^2]/81 = 1

The answer would be:

[±9(x-5)/2] + [-3]; student inserts -3.

[equation_of_a_hyperbola.html?seed=58&problem=0](http://sandcastle.khanacademy.org/media/castles/Khan:master/exercises/equation_of_a_hyperbola.html?seed=58&problem=0&debug)

[["3+9(x-5)/2"]]

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