Aus Online Mathematik Brückenkurs 1

Version vom 09:37, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 4.1:6c moved to Solution 4.1:6c: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 11:45, 27. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
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-	{{NAVCONTENT_START}}	+	What we need to do is to rewrite the equation in the standard form
-	<center> [[Image:4_1_6c-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
		+	<math>\left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}</math>
		+
		+
		+	because then we can read off the circle's centre
		+	<math>\left( a \right.,\left. b \right)</math>
		+	and radius,
		+	<math>r</math>.
		+
		+	In our case, we need only take out the factor
		+	<math>~\text{3}</math>
		+	from the brackets on the left-hand side
		+
		+
		+	<math>\begin{align}
		+	& \left( 3x-1 \right)^{2}+\left( 3y+7 \right)^{2}=3^{2}\left( x-\frac{1}{3} \right)^{2}+3^{2}\left( y+\frac{7}{3} \right)^{2} \\
		+	& =9\left( x-\frac{1}{3} \right)^{2}+9\left( y+\frac{7}{3} \right)^{2} \\
		+	\end{align}</math>
		+
		+
		+	and then divide both sides by
		+	<math>\text{9}</math>
		+	, so as to get the equation in the desired form:
		+
		+
		+	<math>\left( x-\frac{1}{3} \right)^{2}+\left( y+\frac{7}{3} \right)^{2}=\frac{10}{9}</math>
		+
		+	Because the right-hand side can be written as
		+	<math>\left( \sqrt{\frac{10}{9}} \right)^{2}</math>
		+	and the term
		+	<math>\left( y+\frac{7}{3} \right)^{2}</math>
		+	as
		+	<math></math>
		+
		+	<math>\left( y-\left( -\frac{7}{3} \right) \right)^{2}</math>, the equation describes a circle with its centre at
		+	<math>\left( \frac{1}{3} \right.,\left. -\frac{7}{3} \right)</math>
		+	and radius
		+	<math>\sqrt{\frac{10}{9}}=\frac{\sqrt{10}}{3}</math>
		+
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	<center> [[Image:4_1_6c-2(2).gif]] </center>		<center> [[Image:4_1_6c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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Version vom 11:45, 27. Sep. 2008

What we need to do is to rewrite the equation in the standard form

\displaystyle \left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}

because then we can read off the circle's centre \displaystyle \left( a \right.,\left. b \right) and radius, \displaystyle r.

In our case, we need only take out the factor \displaystyle ~\text{3} from the brackets on the left-hand side

\displaystyle \begin{align} & \left( 3x-1 \right)^{2}+\left( 3y+7 \right)^{2}=3^{2}\left( x-\frac{1}{3} \right)^{2}+3^{2}\left( y+\frac{7}{3} \right)^{2} \\ & =9\left( x-\frac{1}{3} \right)^{2}+9\left( y+\frac{7}{3} \right)^{2} \\ \end{align}

and then divide both sides by \displaystyle \text{9} , so as to get the equation in the desired form:

\displaystyle \left( x-\frac{1}{3} \right)^{2}+\left( y+\frac{7}{3} \right)^{2}=\frac{10}{9}

Because the right-hand side can be written as \displaystyle \left( \sqrt{\frac{10}{9}} \right)^{2} and the term \displaystyle \left( y+\frac{7}{3} \right)^{2} as \displaystyle

\displaystyle \left( y-\left( -\frac{7}{3} \right) \right)^{2}, the equation describes a circle with its centre at \displaystyle \left( \frac{1}{3} \right.,\left. -\frac{7}{3} \right) and radius \displaystyle \sqrt{\frac{10}{9}}=\frac{\sqrt{10}}{3}