ellipse Watch

andrewlee89
Badges: 1
Rep:
?
#1
Report Thread starter 10 years ago
#1
Consider the quartic equation \displaystyle Ax^2 + Cy^2 + Dx + Ey + F = 0

If AC> 0 show that it leads to an ellipse.

RIght, if AC > 0, it imples (A and C) < 0 OR (A and C) > 0. After completing the square etc I managed to get an equation of an ellipse but, when I plugged in values for A and C in graphmatica, I found that if (A and C) > 0 (positive values) the graph could not be plotted. Can someone please explain this to me?
0
quote
reply
Mr M
  • Community Assistant
Badges: 20
Rep:
?
#2
Report 10 years ago
#2
(Original post by andrewlee89)
Consider the quartic equation \displaystyle Ax^2 + Cy^2 + Dx + Ey + F = 0

If AC> 0 show that it leads to an ellipse.

RIght, if AC > 0, it imples (A and C) < 0 OR (A and C) > 0. After completing the square etc I managed to get an equation of an ellipse but, when I plugged in values for A and C in graphmatica, I found that if (A and C) > 0 (positive values) the graph could not be plotted. Can someone please explain this to me?
I'm afraid you are wrong.

If A=1, C=2, D=0, E=0, F=-1 you have an ellipse and A and C are both positive so this condition is not preventing the graph from being plotted.

And it isn't a quartic equation.
0
quote
reply
andrewlee89
Badges: 1
Rep:
?
#3
Report Thread starter 10 years ago
#3
how is x^2 + 2y^2 = 0 an ellipse?
0
quote
reply
Hancock orbital
Badges: 1
Rep:
?
#4
Report 10 years ago
#4
Anything in the form ax^2 + by^2 + cy + dx = z is an ellipse unless a = b, in which case you have the special case. The special case is a circle.
0
quote
reply
Mr M
  • Community Assistant
Badges: 20
Rep:
?
#5
Report 10 years ago
#5
(Original post by andrewlee89)
how is x^2 + 2y^2 = 0 an ellipse?
Ok, I modified my previous result to make you feel happier! Your reasoning was incorrect in any event.
0
quote
reply
andrewlee89
Badges: 1
Rep:
?
#6
Report Thread starter 10 years ago
#6
yea well im sorry i got it wrong lol! is the circle the only special case?
0
quote
reply
Davidosh
Badges: 0
Rep:
?
#7
Report 10 years ago
#7
(Original post by Mr M)
Ok, I modified my previous result to make you feel happier! Your reasoning was incorrect in any event.
I've gotanother ellipse question, if a standard ellipse (X/a)^2+(Y/B)^2=1 and is shifted to (h,0) what is the new equation foci and directrices so far I have (x-H/a)^2 + (y/B)^2=1 for the equation is that right? From then on im completely stuck any ideas
0
quote
reply
Hancock orbital
Badges: 1
Rep:
?
#8
Report 10 years ago
#8
That should be (x/a - h)^2 + (y/b)^2 = 1
0
quote
reply
Mr M
  • Community Assistant
Badges: 20
Rep:
?
#9
Report 10 years ago
#9
(Original post by Davidosh)
I've gotanother ellipse question, if a standard ellipse (X/a)^2+(Y/B)^2=1 and is shifted to (h,0) what is the new equation foci and directrices so far I have ((x-h)/a)^2 + (y/B)^2=1 for the equation is that right? From then on im completely stuck any ideas
Hancock's answer was incorrect. Yours was correct (assuming you had a bracket where I added it).

You also need to translate the foci and directices in exactly the same way.

The bottom example on this page will help.

http://www.valleyview.k12.oh.us/vvhs...ipseguide.html
0
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Fri, 14 Dec '18
  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19

Were you ever put in isolation at school?

Yes (114)
26.27%
No (320)
73.73%

Watched Threads

View All