\[\begin{array}{l}
(E):\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 \Rightarrow \left\{ \begin{array}{l}
x = 3\sin t\\
y = 2\cos t
\end{array} \right.,t \in [0;2\pi )\\
\left\{ \begin{array}{l}
A(3; - 2)\\
B( - 3;2)
\end{array} \right. \Rightarrow \overrightarrow {AB} = ( - 6;4)\\
C({x_0};{y_0}) \in (E),{x_0} > 0,{y_0} > 0\\
\Rightarrow \left\{ \begin{array}{l}
{x_0} = 3\sin t > 0\\
{y_0} = 2\cos t > 0
\end{array} \right.,t \in [0;2\pi ) \Rightarrow \left\{ \begin{array}{l}
\sin t > 0\\
\cos t > 0
\end{array} \right.\\
\Rightarrow \overrightarrow {AC} = (3\sin t - 3;2\cos t + 2)\\
{S_{ABC}} = \frac{1}{2}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right| = \frac{1}{2}\left| { - 6(2\cos t + 2) - 4(3\sin t - 3)} \right| = 6\left| {\cos t + \sin t} \right|
\end{array}\]
$S_{ABC}$ đạt GTLN chỉ khi $\cos t + \sin t$ đạt GTLN với $t \in [0;\dfrac{\pi}{2} )$ hay:
\[t = \frac{\pi }{4} \Rightarrow {S_{\max }} = 6\sqrt 2 \Rightarrow \left\{ \begin{array}{l}
{x_0} = \frac{{3\sqrt 2 }}{2}\\
{y_0} = \sqrt 2
\end{array} \right. \Rightarrow C(\frac{{3\sqrt 2 }}{2};\sqrt 2 )\]
Cách thầy mình dạy đấy.
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