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Question

Cho

$x,y$ thỏa:

$\frac{x^2}{9}+\frac{y^2}{16}=10$ .Tìm

$Max,Min$

$P=x-y+2016$ .

★★.P.I.N.O.★★ · Answer 1 · 3/1/2016 9:21:40 PM

Ta có:

$\frac{x^2}{9}+\frac{y^2}{16}=10\Leftrightarrow (\frac{x}{3\sqrt{10}})^2+(\frac{y}{4\sqrt{10}})^2=1$

Đặt

$\sin \alpha =\frac{x}{3\sqrt{10}}; \cos \alpha=\frac{y}{4\sqrt{10}},\alpha \in [0;2\pi]$ ta có:

$\mathbb {P=}3\sqrt{10}\sin \alpha-4\sqrt{10} \cos \alpha +2016$

$=5\sqrt{10}(\frac{3}{5}. \sin \alpha-\frac{4}{5}. \cos \alpha)+2016$

$=5\sqrt{10} \sin(\alpha - \beta)+2016$ (với

$\cos \beta=\frac{3}{5};\sin \beta=\frac{4}{5}$ )

Vì

$-1 \le \sin(\alpha - \beta) \le 1$ nên

$-5\sqrt{10}+2016 \le P \le 5\sqrt{10}+2016.$

$\min P$ đạt được tại:

$x=3\sqrt{10}\sin \alpha;y=4\sqrt{10}\cos \alpha$ với

$\alpha= \beta + \frac{3\pi}{2}$

$\max P$ đạt được tại:

$x=3\sqrt{10}\sin \alpha;y=4\sqrt{10}\cos \alpha$ với

$\alpha= \beta + \frac{\pi}{2}.$

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