Question

Ax^2+Cy^2+F=0
A,C,F don't equal 0
A & C are of the same sign and F is of the opposite sigh
what is the equation of the ellipse when the center (0,0) if A doesn't equal C

Answer 1

Do you know the form of an an equation for an ellipse centered at the origin?

Answer 2

@chinenyeogueri ?

Answer 3

yes x^2/a^2 + y^/b^2 = 1

Answer 4

yup yup - so we can mirror that form


\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]

with

\[ \frac{A}{-F}x^2+\frac{C}{-F}y^2 =1 \]

Remember that you can multiply any term by 1 without changing it, so

\[\frac{A}{-F}x^2 = 1*\frac{A}{-F}x^2= \frac{ \frac{1}{A}}{\frac{1}{A}}\frac{A}{-F}x^2=\frac{x^2}{\frac{-F}{A}}\]
so
\[a^2 = \frac{-F}{A}\]

similar process to find b^2

Sorry that took a while :P

Answer 5

whats the frac thing

Answer 6

hello

Answer 7

The first weird fraction is (1/A) divided by (1/A) , since anything divided by itself is 1. It's to get the A in the denominator to mirror the form of the equation of an ellipse you posted. The second fraction thing is x^2 divided by (-F/A).

\[ \frac{ \ \big(\frac{1}{A}\big) \ }{ \ \big(\frac{1}{A}\big) \ }\frac{A}{-F} x^2=\frac{ \ \big(\frac{A}{A}\big) \ }{ \ \big(\frac{-F}{A}\big) \ } x^2=\frac{ \ 1 \ }{ \ \big(\frac{-F}{A}\big) \ }x^2=\frac{ \ x^2 \ }{ \ \big(\frac{-F}{A}\big) \ }=\frac{x^2}{a^2}
\\ \hspace{4.1in} a^2 =\frac{-F}{A}\]

So for the y term it'd be


\[ \frac{ \ \big(\frac{1}{C}\big) \ }{ \ \big(\frac{1}{C}\big) \ }\frac{C}{-F} y^2=\frac{ \ \big(\frac{C}{C}\big) \ }{ \ \big(\frac{-F}{C}\big) \ } y^2=\frac{ \ 1 \ }{ \ \big(\frac{-F}{C}\big) \ }y^2=\frac{ \ y^2 \ }{ \ \big(\frac{-F}{C}\big) \ }=\frac{y^2}{b^2}
\\ \hspace{4.1in} b^2 =\frac{-F}{C}\]

and the equation of the ellipse would be

\[ \frac{ \ x^2 \ }{ \ \big(\frac{-F}{A}\big) \ } + \frac{ \ y^2 \ }{ \ \big(\frac{-F}{C}\big) \ }=1\]