Giả sử $M(a;b)\in (E)$
$T_{\overrightarrow{u}}: (E)\to (E')$
$\to M'(x;y)$
Ta có: $x=a-2; y=b+4$
$\to a=x+2; b=y-4$
$\to M(x+2; y-4)$
$M\in (E)$ nên ta có:
$\dfrac{(x+2)^2}{16}+\dfrac{(y-4)^2}{9}=1$
$\to \dfrac{x^2+4x+4}{16}+\dfrac{y^2-8y+16}{9}=1$
$\to 9(x^2+4x+4)+16(y^2-8y+16)=0$
$\to 9x^2+16y^2+36x-128y+292=0$
Vậy phương trình elip $(E')$ là:
$(E'): 9x^2+16y^2+36x-128y+292=0$