I forgot one question-in a hyperbola, the "a^2" is always under the first variable, whether it be "x" or "y", correct
2. and 3.
I always associate a^2 with the x's
and b^2 with the y's, but that is a personal choice.
It will depend how your text or your instructor chooses.
regarding your comment, ...
what if I simply changed the equation to
(x-6)^2/72 - (y-4)^2 /49 = -1 , (which is the preferred method of writing it)
would you now label a^2 as 72 ?
Can you see the problem ?
That is why associating a^2 with the x's and b^2 with the y's makes so much more sense.
4. c^2 = a^2 + b^2 = 121
c = 11
foci : (6, 15) and (6, -7)
(your formula for a vertical hyperbola is correct)
So if I change the formula to = (-1), then I have to change the a^2 to be under the "x" is that correct?
Thank you for your help
All I did was to multiply each term by -1
I like to have all my conic section equations in standard form, that is ,
my circle, ellipse and hyperbola equations start with the "x" term.
As I said, I associate the a^2 with the x's and the b^2 with the y's.
This saves a lot of confusion.
However, if your instructor wants you to do it otherwise, such as the a's always going with the major axis, then do so.
okay-thank you I got it now
Answer this Question
Algebra II-Please check - I'm sorry the computer jammed on an earlier post and I...
Algebra - HELP! I am not sure how I would even start this problem or solve it. A...
College Algebra - Can we try this again? I already posted this question and ...
Trig - Find an equation for the hyperbola described. Foci at (-4,0) and (4,0); ...
math30 - The equation of the hyperbola is (x-3)^2/4 - (y+1)^2/16 =-1. What is ...
Math--Please Check!!! - Hi, I asked this question yesterday and I was given help...
Algebra - I also need you to check and make sure that I am using the FOIL method...
to bobpursley,math - This is what i have am i wrong or correct.Or how can word ...
algebra - i have two answers to this problem so am not sure which way I need to ...
Algebra Zero Product Rule - I'm working with the zero product rule and I've ...