Deriving the Standard form for an ellipse from its Quadratic form

[Dale10101]

The attached docs show my effort which I believe is accurate because I made numerical checks along the way. The result however is rather complex. I understand the coefficients in the two forms are not equivalent ("a" in the ellipse standard form is not the same as the "a" in the quadratic standard form etc). Even so I am thinking that there might be a cleaner result ... but maybe not or maybe I made an error.

BTW using computer aided software to do basic algebra is seldom efficient, the software usually wants to simplify after performing operations and so prevents one from achieving the desired result. I much prefer pencil and paper BUT, CAS has the BIG advantage of enabling one to numerically verify results step by step and prevent the usual stupid addition errors etc. Have not found any package that makes simple manual operations easier. Sheesh!

If you jump to the bottom of the second page ... the question is, can the result be made simpler?

 

[soroban]

Hello, Dale10101!

The standard form of an ellipse is: .\(\displaystyle \dfrac{(x-H)^2}{A^2} + \dfrac{(y-K)^2}{B^2} \:=\:1\)


I agree with #22: .\(\displaystyle a\left(x + \dfrac{b}{2a}\right)^2 + c\left(y + \dfrac{d}{2c}\right)^2 \:=\:\dfrac{ad^2 + cb^2 - 4ace}{4ac}\)

But standard form goes like this . . .


Divide by \(\displaystyle \dfrac{ad^2+cb^2-4ace}{4ac}\!:\)


. . \(\displaystyle \dfrac{a\left(x+\dfrac{b}{2a}\right)^2}{\;\dfrac{ad^2+cb^2-4ace}{4ac}\;} + \dfrac{c\left(y+\dfrac{d}{2c}\right)^2}{\;\dfrac{ad^2 + cb^2-4ace}{4ac}\;} \;=\;1\)


. . \(\displaystyle \dfrac{\left(x+\dfrac{b}{2a}\right)^2}{\;\dfrac{ad^2+cb^2-4ace}{4a^2c}\;} + \dfrac{\left(y+\dfrac{d}{2c}\right)^2}{\;\dfrac{ad^2+cb^2-4ace}{4ac^2}\;} \;=\;1 \)

 

[Dale10101]

got it

Ok. great. And because h,k, a & b in the standard formula for the ellipse have no necessary relationship to one another that last is the simplest final form. Good deal. Thanks.