4. The numbers x and y are solutions to the system $\left\{\begin{array}{l}a x-y=2 a+1 \\ -x+a y=a\end{array}\right.$, where a is a parameter. What is the smallest value that the expression $2 y^{2}$ $x^{2}$ can take if $a \in[-0.5 ; 2] .(20$ points)
Solution: $\left\{\begin{array}{l}a x-y=2 a+1 \\ x=a(y-1)\end{array} \Leftrightarrow\left\{\begin{array}{l}a^{2}(y-1)-y=2 a+1 \\ x=a(y-1)\end{array} \Leftrightarrow\left\{\begin{array}{l}y\left(a^{2}-1\right)=a^{2}+2 a+1 \\ a y-a=x\end{array}\right.\right.\right.$

For $a \neq \pm 1$

$\left\{\begin{array}{l}y=\frac{a+1}{a-1} \\ x=\frac{2 a}{a-1}\end{array}\right.$

$F(a)=2 y^{2}-x^{2}=2\left(\frac{a+1}{a-1}\right)^{2}-\left(\frac{2 a}{a-1}\right)^{2}=\frac{2\left(a^{2}+2 a+1-2 a^{2}\right)}{(a-1)^{2}}=\frac{-2\left(a^{2}-2 a-1\right)}{(a-1)^{2}}=-2+\frac{4}{(a-1)^{2}}$

$F(-0.5)=-2 / 9$.

Since the fraction is always positive, the smallest value it takes is when $(a-1)$ becomes the largest in absolute value, i.e., when $a = 0.5$.

Let's find the value of the expression when $\mathrm{a}= \pm 1$.

When $\mathrm{a}=-1$, the system has infinitely many solutions, and when $\mathrm{a}=1$, the system has no solutions.

Answer: $\mathrm{a}=-0.5$, the minimum value is $-2 / 9$.