Algebra II-Please check fpr explanation
posted by Graham .
I have a question concerning this problem:I just want to make sure my formulas are correct before I do problem
It is a hyperbola equation
(y-4)^2/49 - (x-6)^2/72 = 1
1.It is a vertical hyperbola, correct?
2. the a^2 is 49, correct?
3. the b^2 is 72, correct?
4. I find the foci by a^2 + b^2 = c^2, and when I have the square root of "c^2" it would be ( "h" which is 6 in this case, k which is 4 +-whatever "c" ended up to be, correct?
5.Vertices for vertical hyperbola is (h,k+-a), correct
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