Trong mặt phẳng tọa độ Oxy cho elip có ohương trình chính tắt $\frac{x^{2}}{16} + \frac{y^{2}}{9} =1$
và 2 điểm A(4;-3) B(-4;3). Tìm điểm C thuộc E sao cho diện tích tam giác ABC lớn nhất

* AB = 10. Phương trình đường thẳng AB: 3x + 4y = 0.

* Do $C \in (E)$, ta gọi  $C(4\cos t,3\sin t),t \in \left[ { - \pi ;\pi } \right]$

* ${S_{ABC}} = \frac{1}{2}AB.d(C,AB)$

Suy ra ${S_{ABC}}$ lớn nhất $\Leftrightarrow d(C,AB)$ lớn nhất.

* $d(C,AB) = \frac{{\left| {12\cos t + 12\sin t} \right|}}{5} = \frac{{12\sqrt 2 \left| {\sin (t + \frac{\pi }{4})} \right|}}{5}$

Suy ra: $Maxd(C,AB) = \frac{{12\sqrt 2 }}{5} \Leftrightarrow \sin (t + \frac{\pi }{4}) =  \pm 1 \Leftrightarrow t = \frac{\pi }{4} \vee t =  - \frac{{3\pi }}{4} \Leftrightarrow C(\frac{{4\sqrt 2 }}{2};\frac{{3\sqrt 2 }}{2}) \vee C(\frac{{ - 4\sqrt 2 }}{2};\frac{{ - 3\sqrt 2 }}{2})$

Vậy \[Max{S_{ABC}} = 12\sqrt 2  \Leftrightarrow C(\frac{{4\sqrt 2 }}{2};\frac{{3\sqrt 2 }}{2}) \vee C(\frac{{ - 4\sqrt 2 }}{2};\frac{{ - 3\sqrt 2 }}{2})\]

Ừ, mình sửa lại rồi. –  binhnguyenhoangvu 09-05-13 04:00 PM
nếu giải theo hướng này thì max d(C;AB) khi |sin(t pi/4)| max chứ bạn, còn thiếu 1 nghiệm t nữa ^^ –  shinigem_1994 09-05-13 03:55 PM

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