Goi $M(a,b) $ thuoc (E) khi do ta co
$a^{2}+4b^{2}=8$
Theo BDT Bunhia co
$(3a+4b)^{2}\leq (3^{2}+2^{2})(a^{2}+(2b)^{2})=104$
$<=>-2\sqrt{26}\leq 3a+4b\leq 2\sqrt{26}$
$<=>-2\sqrt{26}+24\leq 3a+4b+24\leq 2\sqrt{26}+24$
$<=>-2\sqrt{26}+24\leq \left| {3a+4b+24} \right|\leq 2\sqrt{26}+24$
$<=>\frac{-2\sqrt{26}+24}{5}\leq \frac{\left| {3a+4b+24} \right|}{5}\leq \frac{2\sqrt{26}+24}{5}$
$<=>\frac{-2\sqrt{26}+24}{5}\leq d(M;d)\leq \frac{2\sqrt{26}+24}{5}$
Vay $d(M;d)min=\frac{-2\sqrt{26}+24}{5}$
Tim M.....