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| Title of object: |
homogeneous equation |
| Canonical Name: |
HomogeneousEquation |
| Type: |
Topic |
| Created on: |
2005-05-07 14:29:41 |
| Modified on: |
2005-05-07 14:29:41 |
| Classification: |
msc:00A99, msc:26B35, msc:26C05 |
| Keywords: |
proportional |
Preamble:
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\usepackage{amssymb}
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\newtheorem*{thmplain}{Theorem} |
Content:
The {\em homogeneous equation}
$$f(x,\,y) = 0,$$
where the left hand \PMlinkescapetext{side} is a homogeneous polynomial of degree $r$ in $x$ and $y$, \,determines the ratio $x/y$ between the indeterminates. \,One can be persuaded of this by dividing both \PMlinkescapetext{sides} of the equation by $y^r$. \,Then the left \PMlinkescapetext{side} depends only on $x/y$ (which may be denoted e.g. by $t$).
\textbf{Examples}
\begin{itemize}
\item The equation \,$5x+8y = 0$\, determines that \,$x/y = -\frac{8}{5}$.
\item The equation \,$x^2-7xy+10y^2$\, determines that \,$x/y = 2$\, or \,$x/y = 5$\, (these values are obtained by first dividing both \PMlinkescapetext{sides} of the equation by $y^2$ and then solving the equation \,$(x/y)^2-7(x/y)+10 = 0$).
\item The equation \,$2x^3-x^2y-6xy^2+3 = 0$\, determines that
\,$x/y = \frac{1}{2}$\, or \,$x/y = \pm\sqrt{3}$ (first divide the equation by $y^3$ and then solve \,$2(x/y)^3-(x/y)^2-6(x/y)+3 = 0$).
\end{itemize} |
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