Weise nach, dass $S_{a}S\_\{a\}$ für jeden Wert von $aa$ auf der $yy$-Achse liegt

Tangentengleichungen

1. Die Tangente $t_{f_a}t\_\{f\_\{a\}\}$ an den Graphen von $f_{a}f\_\{a\}$ im Punkt $\left(\frac{5}{a}\mid f_a\left(\frac{5}{a}\right)\right)\backslash left(\backslash frac\{5\}\{a\}\; \backslash left\backslash lvert\backslash ,\; f\_\{a\}\backslash left(\backslash frac\{5\}\{a\}\backslash right)\backslash right.\backslash right)$ hat die Steigung $\frac{1}{a^2}\backslash frac\{1\}\{a^\{2\}\}$.

Berechne $f_a\left(\frac{5}{a}\right)=-{\displaystyle \frac{a}{250}}\cdot {\left({\displaystyle \frac{5}{a}}\right)}^4+{\displaystyle \frac{1}{25}}\cdot {\left({\displaystyle \frac{5}{a}}\right)}^3={\displaystyle \frac{\mathrm{2,5}}{a^3}}f\_\{a\}\backslash left(\backslash frac\{5\}\{a\}\backslash right)=-\backslash dfrac\{a\}\{250\}\backslash cdot\backslash left(\backslash dfrac\{5\}\{a\}\backslash right)^4+\backslash dfrac\{1\}\{25\}\backslash cdot\backslash left(\backslash dfrac\{5\}\{a\}\backslash right)^3=\backslash dfrac\{2\{,\}5\}\{a^3\}$.

Dann ist $t_{f_a}:y={\displaystyle \frac{1}{a^2}}\cdot x+bt\_\{f\_\{a\}\}:\; y=\backslash dfrac\{1\}\{a^2\}\backslash cdot\; x+b$.

Setze den Punkt $({\displaystyle \frac{5}{a}}\mid {\displaystyle \frac{\mathrm{2,5}}{a^3}})\backslash left(\backslash dfrac\{5\}\{a\}\backslash mid\; \backslash dfrac\{2\{,\}5\}\{a^3\}\backslash right)$ ein:

${\displaystyle \frac{\mathrm{2,5}}{a^3}}={\displaystyle \frac{1}{a^2}}\cdot {\displaystyle \frac{5}{a}}+b\Rightarrow b=-{\displaystyle \frac{\mathrm{2,5}}{a^3}}\backslash dfrac\{2\{,\}5\}\{a^3\}=\backslash dfrac\{1\}\{a^2\}\backslash cdot\; \backslash dfrac\{5\}\{a\}+b\backslash ;\backslash Rightarrow\backslash ;\; b=-\backslash dfrac\{2\{,\}5\}\{a^3\}$

Die Tangentengleichung lautet: $t_{f_a}:y={\displaystyle \frac{1}{a^2}}\cdot x-{\displaystyle \frac{\mathrm{2,5}}{a^3}}t\_\{f\_\{a\}\}:\; y=\backslash dfrac\{1\}\{a^2\}\backslash cdot\; x-\backslash dfrac\{2\{,\}5\}\{a^3\}$.

2. Die Tangente $t_{g_a}t\_\{g\_\{a\}\}$ an den Graphen von $g_{a}g\_\{a\}$ im Punkt $\left(\frac{5}{a}\mid g_a\left(\frac{5}{a}\right)\right)\backslash left(\backslash frac\{5\}\{a\}\; \backslash left\backslash lvert\backslash ,\; g\_\{a\}\backslash left(\backslash frac\{5\}\{a\}\backslash right)\backslash right.\backslash right)$ hat die Steigung ${\displaystyle \frac{5-3a^2}{5a^2}}={\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}}\backslash dfrac\{5-3\; a^\{2\}\}\{5\; a^\{2\}\}=\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}$.

Berechne $g_a\left(\frac{5}{a}\right)=f_a\left(\frac{5}{a}\right)-{\displaystyle \frac{3}{5}}\cdot {\displaystyle \frac{5}{a}}={\displaystyle \frac{\mathrm{2,5}}{a^3}}-{\displaystyle \frac{3}{a}}g\_\{a\}\backslash left(\backslash frac\{5\}\{a\}\backslash right)=f\_\{a\}\backslash left(\backslash frac\{5\}\{a\}\backslash right)-\backslash dfrac\{3\}\{5\}\backslash cdot\; \backslash dfrac\{5\}\{a\}=\backslash dfrac\{2\{,\}5\}\{a^3\}-\backslash dfrac\{3\}\{a\}$.

Dann ist $t_{g_a}:y=({\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}})\cdot x+bt\_\{g\_\{a\}\}:\; y=\backslash left(\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}\backslash right)\backslash cdot\; x+b$.

Setze den Punkt $({\displaystyle \frac{5}{a}}\mid {\displaystyle \frac{\mathrm{2,5}}{a^3}}-{\displaystyle \frac{3}{a}})\backslash left(\backslash dfrac\{5\}\{a\}\backslash mid\; \backslash dfrac\{2\{,\}5\}\{a^3\}-\backslash dfrac\{3\}\{a\}\backslash right)$ ein:

${\displaystyle \frac{\mathrm{2,5}}{a^3}}-{\displaystyle \frac{3}{a}}=({\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}})\cdot \left({\displaystyle \frac{5}{a}}\right)+b={\displaystyle \frac{5}{a^3}}-{\displaystyle \frac{3}{a}}+b\Rightarrow b=-{\displaystyle \frac{\mathrm{2,5}}{a^3}}\backslash dfrac\{2\{,\}5\}\{a^3\}-\backslash dfrac\{3\}\{a\}=\backslash left(\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}\backslash right)\backslash cdot\; \backslash left(\backslash dfrac\{5\}\{a\}\backslash right)+b=\backslash dfrac\{5\}\{a^3\}-\backslash dfrac\{3\}\{a\}+b\backslash ;\backslash Rightarrow\backslash ;b=-\backslash dfrac\{2\{,\}5\}\{a^3\}$

Die Tangentengleichung lautet: $t_{g_a}:y=({\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}})\cdot x-{\displaystyle \frac{\mathrm{2,5}}{a^3}}t\_\{g\_\{a\}\}:\; y=\backslash left(\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}\backslash right)\backslash cdot\; x-\backslash dfrac\{2\{,\}5\}\{a^3\}$.

Schnittpunkt

Beide Tangentengleichungen haben den gleichen y-Achsenabschnitt, d.h. sie schneiden sich im Punkt $S_a(0\mid -{\displaystyle \frac{\mathrm{2,5}}{a^3}})S\_a\backslash left(0\backslash mid\; -\backslash dfrac\{2\{,\}5\}\{a^3\}\backslash right)$.

$S_{a}S\_\{a\}$ liegt für jeden Wert von $aa$ auf der y-Achse.

Untersuche, für welche Werte von $a\in \mathbb{R}a\; \backslash in\; \backslash mathbb\{R\}$ mit $a>0a>0$ das Dreieck $S_a{G}_a{F}_{a}S\_\{a\}\; G\_\{a\}\; F\_\{a\}$ rechtwinklig ist

Die Tangentengleichung lautet $t_{f_a}:y={\displaystyle \frac{1}{a^2}}\cdot x-{\displaystyle \frac{\mathrm{2,5}}{a^3}}t\_\{f\_\{a\}\}:\; y=\backslash dfrac\{1\}\{a^2\}\backslash cdot\; x-\backslash dfrac\{2\{,\}5\}\{a^3\}$.

Die Steigung dieser Tangente ist ${\displaystyle \frac{1}{a^2}}\backslash dfrac\{1\}\{a^2\}$und für $a>a>$0 ist die Steigung größer null.

Die Gerade mit der Gleichung $x=\frac{5}{a}x=\backslash frac\{5\}\{a\}$ steht senkrecht zur x-Achse. Deshalb kann sich im Punkt $F_{a}F\_a$ kein rechter Winkel befinden.

Der rechte Winkel kann nur im Punkt $G_{a}G\_a$ sein.

Die Tangentengleichung lautet: $t_{g_a}:y=({\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}})\cdot x-{\displaystyle \frac{\mathrm{2,5}}{a^3}}t\_\{g\_\{a\}\}:\; y=\backslash left(\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}\backslash right)\backslash cdot\; x-\backslash dfrac\{2\{,\}5\}\{a^3\}$.

Die Steigung dieser Tangente ist ${\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}}\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}$.

Setze diese Steigung gleich null und löse nach $aa$ auf:

${\displaystyle \frac{1}{a^2}}-{\displaystyle \frac{3}{5}}=0\Rightarrow a^2={\displaystyle \frac{5}{3}}\Rightarrow a_{\mathrm{1,2}}=\pm \sqrt{{\displaystyle \frac{5}{3}}}\backslash dfrac\{1\}\{a^2\}-\backslash dfrac\{3\}\{5\}=0\backslash ;\backslash Rightarrow\backslash ;a^2=\backslash dfrac\{5\}\{3\}\backslash ;\backslash Rightarrow\backslash ;a\_\{1\{,\}2\}=\backslash pm\backslash sqrt\{\backslash dfrac\{5\}\{3\}\}$

Wegen $a>0a>0$ erhält man die Lösung $a=\sqrt{{\displaystyle \frac{5}{3}}}a=\backslash sqrt\{\backslash dfrac\{5\}\{3\}\}$.

Für diesen Wert von $aa$ ist das Dreieck $S_a{G}_a{F}_{a}S\_\{a\}\; G\_\{a\}\; F\_\{a\}$ rechtwinklig.