• Jan 16th 2009, 07:05 PM
mathproject
can i please get some help in this one. trying to figure it out and i have atest tomm....
Five of the six players on the receiving team must stand in an approximately elliptical arrangement, as shown in the diagram below.
1. If point A is the origin of the coordinate axes, state the standard
form equation

for this ellipse.
2. If point B is the origin of the coordinate axes, determine the generalform equation for the ellipse
• Jan 16th 2009, 07:10 PM
Jhevon
Quote:

Originally Posted by mathproject
can i please get some help in this one. trying to figure it out and i have atest tomm....
Five of the six players on the receiving team must stand in an approximately elliptical arrangement, as shown in the diagram below.
1. If point A is the origin of the coordinate axes, state the standard
form equation

for this ellipse.
2. If point B is the origin of the coordinate axes, determine the generalform equation for the ellipse

see post #2 here, under the heading "Ellipse"

of course, you are expected to know what phrases like "major axis", "minor axis" etc mean

if you can find the center and the major and minor axis of the ellipse with respect to A and then B, then you are set

can you continue?
• Jan 16th 2009, 07:25 PM
mathproject
is it right?
well here is the answer for A. i guess
(x-3)^2/9 + 4(y-9/2)^2/81=1

is it right?
still strugling with the second part B(Worried)
• Jan 16th 2009, 07:32 PM
galactus
That is correct for #1.

Where is the center for #2?. It's at (-6,0) is it not?. But the lengths of the axes have not changed.
• Jan 16th 2009, 07:57 PM
mathproject
i try to use(-6,0) but i get stuck wen it comes to 4(y-9/2)^2/81=1 . and then evrything is relly confusing for me. can some1 explain plz
• Jan 16th 2009, 08:06 PM
Jhevon
Quote:

Originally Posted by mathproject
i try to use(-6,0) but i get stuck wen it comes to 4(y-9/2)^2/81=1 . and then evrything is relly confusing for me. can some1 explain plz

if B is the origin, (-6,0) is the center. as before, the major axis is the vertical one, it is of length 9, and the minor axis is of length 6

thus, we have

(h,k) = (-6,0)
a = 9/2
b = 3

now simply plug this into the form as the thread i directed you to directs

we have:

$\displaystyle \frac {(x + 6)^2}{3^2} + \frac {y^2}{(9/2)^2} = 1$
• Jan 16th 2009, 08:31 PM
mathproject
Quote:

Originally Posted by Jhevon
if B is the origin, (-6,0) is the center. as before, the major axis is the vertical one, it is of length 9, and the minor axis is of length 6

thus, we have

(h,k) = (-6,0)
a = 9/2
b = 3

now simply plug this into the form as the thread i directed you to directs

we have:

$\displaystyle \frac {(x + 6)^2}{3^2} + \frac {y^2}{(9/2)^2} = 1$

ok i expanded the forms 9/2x^2+9y^2+54x+40.5y+12=0

is this correct?
• Jan 16th 2009, 08:56 PM
Jhevon
Quote:

Originally Posted by mathproject
ok i expanded the forms 9/2x^2+9y^2+54x+40.5y+12=0

well i suck at math. i think u should knoe that by now . so is this correct???

no, that's not correct

note that

$\displaystyle \frac {(x + 6)^2}9 + \frac {4y^2}{81} = 1 \implies 81(x + 6)^2 + 36y^2= 729$ .........(i multiplied by 9(81))

now i told you how to expand something like $\displaystyle (x + 6)^2$ already