Question

Its a complex question..
Whe have an ellipse that crosses trough the next 5 points:
A(0,4)
B(6,2)
C(2,-2)
D(-2,-2)
E(-6,2)

Find the ellipse formula

Answer 1

It has many solutions. I will let you know why.

Answer 2

The general equations of a conic is of the form
\[
g(x,y)=a x^2+b x y+c y^2+d x+e y+f=0
\]
You have to find six unknowns and you have five information, you can take one unknown arbitrary. I took f=112 to get nice answer.

Answer 3

The five equations are obtained by putting g (each point) =0
You will get five equations
\[
16 c+4 e+112=0\\
36 a+12 b+4 c+6 d+2 e+112=0\\
4 a-4 b+4 c+2 d-2e+112=0\\
4 a+4 b+4 c-2 d-2 e+112=0
\\ 36 a-12 b+4 c-6 d+2 e+112=0

\]
You put them in wolframalpha and you solve

Answer 4

Actually, I used Mathematica to solve
You find
{a= -3, b= 0, c = -13, d= 0, e= 24}

Answer 5

Your equation will be
\[
-3 x^2-13 y^2+24 y+112=0

\]

Answer 6

Here is the ellipse with your points on it
@ganeshie8 @Loser66 @ikram002p

Answer 7

yess.. just wondering.. if we limit to ellipses whose major/minor axis are parallel to xy axes, then we may get an unique solution...

Answer 8

like the one in ur graph :)

Answer 9

oh its the oly ellipse possible if we limit to translation...
all other ellipses wud require rotation !

Answer 10

Actually, @ganeshie8 you might be right. Try it

Answer 11

Actually, I tried f=2 and I go the same ellipse.

Answer 12

it said in the qs its ellipse so i guss its oly onr sol ,
we just need to match the hypo. wid ellipse fomulla

Answer 13

we cud use the points
B(6,2) ,E(-6,2)
C(2,-2), D(-2,-2)
such that
f'(6)=-f'(-6)
f'(2)=-f'(-2)
hummm but im not sure it wont work :|

Answer 14

i think ya its complex but it can be solved

Answer 15

if u cud solve the equation 1,2,3,4,5

Answer 16

actually, i was searching for the method. The easy way out would have been to just use mathematica or geogebra.

Answer 17

just see eliasab sol !