bai nay kho the

Question

cho E:

$\frac{x^{2}}{8}+\frac{y^{2}}{2}$ =1va duong thang d:3x+4y+24=0
tim M tren E sao cho khoang kach tu M den d min

thiensugacoi_95 · Answer 1 · 6/1/2013 12:11:48 PM

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.....

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