# ellipse equation from 6 points?Watch

#1
Hey, i was just wondering this - (as i recently tried to answer a question on here about how to sketch a graph, and did my own investigation - as ive seen this question twice on here now - and i believe ive kicked its ass! (in a pdf)).

My question is (havent tried it yet):

can you get the cartesion equation of an ellipse if you know 2 points where it intesects both the x and y axes, and the line y=x?

As ive never tried this yet (may be about to), i dont know if the answer is a "simple" yes!

(points being: (BEING EDITED)

(this is only ONE set of points for one ellipse generated by a version of the equation (implicit) that i was investigating, and together the both ellipses make up the "plot")
0
#2
Bump! new info above.
0
4 years ago
#3
(Original post by Hasufel)
Bump! new info above.
an ellipse is a quadratic whose general form is

ax2+by2+cxy+dx+ey+f=0

or dividing by a

x2+Py2+Qxy+Rx+Sy+T=0

So I will say to "pin" 5 constants you need 5 equations(points)
0
4 years ago
#4
(Original post by Hasufel)
Hey, i was just wondering this - (as i recently tried to answer a question on here about how to sketch a graph, and did my own investigation - as ive seen this question twice on here now - and i believe ive kicked its ass! (in a pdf)).

My question is (havent tried it yet):

can you get the cartesion equation of an ellipse if you know 2 points where it intesects both the x and y axes, and the line y=x?

As ive never tried this yet (may be about to), i dont know if the answer is a "simple" yes!

Well the ellipse does not necessarily intersect y=x, or the x axis or the y-axis. e.g. circle radius 1 centered (100,-100)

However, if by some chance it does then you can easily prove it is unique given 2 points where it intesects both the x and y axes, and the line y=x? (exercise...). Getting the Cartesian equation is messy though. Why do you want to use Cartesians though, they are super messy for rotated ellipses. Some parametric system should be fine.

If it is not rotated you only need 4 points (can be any 4) to determine what the equation is (easy).

For rotated you need five points (any I believe)

You will need to solve five simultaneous equations in five variables (in theory there is a quicker algorithm for this sort of thing though - i cannot remember its name):

Use the parametric form given in the link:
http://en.wikipedia.org/wiki/Ellipse...arametric_form

Also, don't do it by hand MATLAB is your friend!
1
#5
gheesh! - i never thought of parametrics! - doi!

off to watch the (English) football! - (the Scottish stuff we get up here is like Aldi cola.....(need i say more?) - so, off to watch some GOOD football!

(they torture us up here by showing it late, forcing us to sit through the Scottish rubbish - which no one without a tatoo watches!)
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

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

### University open days

• Cranfield University
Cranfield Forensic MSc Programme Open Day Postgraduate
Thu, 25 Apr '19
• University of the Arts London
Open day: MA Footwear and MA Fashion Artefact Postgraduate
Thu, 25 Apr '19
• Cardiff Metropolitan University
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (219)
39.32%
No - but I will (38)
6.82%
No - I don't want to (37)
6.64%
No - I can't vote (<18, not in UK, etc) (263)
47.22%