Trong mặt phẳng tọa độ Oxy, cho vecto $\vec{v}=(1,-2) $và đường cong (C): $2x^2+4y^2=1$. Tìm ảnh của (C) qua phép tịnh tiến $T_{\vec{v}}$.

2 câu trả lời

sử dụng quy tích điểm 

`M(x;y)\in(C):T(M)=M'(x':y')in(C)`

`=>{(x'=x+1),(y'=y-2):}`

`=>{(x=x'-1),(y=y'+2):}`

Thay vào `(C)`, ta được 

`2(x+1)^2+4(y+2)^2=1`

`<=>2x^2+4y^2+4x-16y+17=0`

Đáp án: Anh của C qua phép tịnh tiến $T_{\vec{v}}$ là:

$2(x+1)^2+4(y-2)^2=1 \Leftrightarrow 2x^2+4y^2+4x-16y+17+0$

Lời giải:

Câu hỏi trong lớp Xem thêm

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