cứu emmmmmmmmmmmmmmmmmm

Đáp án:

$\begin{array}{l}
a)A\left( {2; - 2} \right) \in \left( P \right)\\
 \Leftrightarrow  - 2 = a{.2^2}\\
 \Leftrightarrow  - 2 = a.4\\
 \Leftrightarrow a =  - \dfrac{1}{2}\\
Vậy\,a =  - \dfrac{1}{2}\\
b)M\left( {x;y} \right) \in y =  - \dfrac{1}{2}{x^2}\\
 \Leftrightarrow M\left( {x; - \dfrac{1}{2}{x^2}} \right)\\
Do:{x_M} - {y_M} = 4\\
 \Leftrightarrow x - \left( { - \dfrac{1}{2}{x^2}} \right) = 4\\
 \Leftrightarrow x + \dfrac{1}{2}{x^2} = 4\\
 \Leftrightarrow 2x + {x^2} = 8\\
 \Leftrightarrow {x^2} + 2x - 8 = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 4 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2 \Leftrightarrow y =  - \dfrac{1}{2}{x^2} =  - 2\\
x =  - 4 \Leftrightarrow y =  - \dfrac{1}{2}{x^2} =  - 8
\end{array} \right.\\
Vậy\,M\left( {2; - 2} \right);M\left( { - 4; - 8} \right)\\
c)M\left( {x; - \dfrac{1}{2}{x^2}} \right)\\
{y_M} = 2{x_M}\\
 \Leftrightarrow  - \dfrac{1}{2}{x^2} = 2.x\\
 \Leftrightarrow  - {x^2} = 4x\\
 \Leftrightarrow {x^2} + 4x = 0\\
 \Leftrightarrow x\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0 \Leftrightarrow y = 0\\
x =  - 4 \Leftrightarrow y =  - 8
\end{array} \right.\\
Vậy\,M\left( {0;0} \right);M\left( { - 4; - 8} \right)
\end{array}$