We had a take home quiz for math on conics and such and I got a bit confused on the matching. I did this part in class looking at my notes, and I thought I was doing it right but then it started looking as if I wasn't. Would you mind checking it? (The bottom six questions)

Click
(The parts on the right (or bottom, I guess) that are cut off just say "= 1")

I changed some of the answers around:
7) A
8) E
9) B
10) D
11) C
12) F

Thanks.

[quote author=hismajesty[yL] link=topic=11473.msg110874#msg110874 date=1115083975]
We had a take home quiz for math on conics and such and I got a bit confused on the matching. I did this part in class looking at my notes, and I thought I was doing it right but then it started looking as if I wasn't. Would you mind checking it? (The bottom six questions)

Click
(The parts on the right (or bottom, I guess) that are cut off just say "= 1")

I changed some of the answers around:
7) A
8) E
9) B
10) D
11) C
12) F

Thanks.
[/quote]

Let's see here
an elipse has the form (x^2)/(a^2)+(y^2)/(b^2)=1
The major axis is along the term whose denominator is larger since if we assume
b > a and x = 0
(y^2)/(b^2)=1 => y=b b > a (major axis is along y axis in this case)
and likewise when y = 0 etc ...
Now its translation is as simple as moving a single point.  Suppose we wish to move the ellipse to the right h units.  Take a point such as (0,b)
Then to move it to (h,b) and have the equation fit the original point I need to subtract h from the desired x coordinate to give me (0,b)...essentially it is (h-h,b).
hence ((x-h)^2)/(a^2)+(y^2)/(b^2)=1 and (h,b) is a point on that ellipse
By the same reasoning you will find that for an ellipse with center (h,k) the form is ((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1
So for the three ellipses we have
A) 9 -> Since the major axis is along the x axis since 25 is larger than 16 and the center is (4,-5) [Hence, (x-4)^2 and (y+5)^2 terms]
B) 7 -> Since the major axis is again along the x axis and the center is (5,4)
C) 11 -> Since the major axis is now along the y axis and the center is (5,-4)

Hyperbolas
A hyperbola (not the rectangular hyperbola) has the form (x^2)/(a^2)-(y^2)/(b^2)=1 (Opens sideways) or (y^2)/(b^2)-(x^2)/(a^2)=1 (Opens vertically)
So to show each one opens only one direction
Assume for example y=0 for the "sideways" one
then (x^2)/(a^2)=1 has a solution x = a
however if we assume a fixed value x, say 0
we get (y^2)/(b^2)=-1 which has no real solution
Use the same reasoning to show this for the vertical one

By similar reasoning replacing the x's and y's with x-h and y-k gives you a hyperbola with center (h,k)
So for our 3 hyperbolas we have
D) 10 -> Since it opens "sideways" since the x term is first and has center (-5,-4) [Again, hence the (x+5)^2 and (y+4)^2 terms]
E) 8 -> Since it opens vertically and has center (5,-4)
F) 12 -> Since it opens "sidrways" and has center (4,5)

Hope this explains a lot.  By the way, I'll be typing a document of tricks for people seeking help and for the tutors ... I work at the math help center at the university and sometimes there are some subjects that I or some of the tutors are not totally familiar with (so we have to waste time looking it up), so this will be easy reference for both tutor and struggling student :P 

I'll post it when I finish it.

Thanks, seems I just had 7 and 9 backwards.