Conics, etc. help

[hismajesty]

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.

[nslay]

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
E
9) B
10) D
11) C
12) F

Thanks.

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  

I'll post it when I finish it.

[hismajesty]

Thanks, seems I just had 7 and 9 backwards.