Giải trâu bò:
Gọi \(N\left(x;y\right)\Rightarrow\left(x-1\right)^2+\left(y+4\right)^2=25\)
\(\left\{{}\begin{matrix}\overrightarrow{NP}=\left(5-x;10-y\right)\\2\overrightarrow{NQ}=\left(16-2x;2-2y\right)\end{matrix}\right.\) \(\Rightarrow\overrightarrow{NP}+2\overrightarrow{NQ}=3.\left(7-x;4-y\right)\)
\(\Rightarrow A=\left|\overrightarrow{NP}+2\overrightarrow{NQ}\right|=3\sqrt{\left(x-7\right)^2+\left(y-4\right)^2}\)
\(A_{min}\) khi \(B=\left(x-7\right)^2+\left(y-4\right)^2\) đạt min
Lượng giác hóa:
Từ giả thiết \(\left(x-1\right)^2+\left(y+4\right)^2=25\Rightarrow\left(\frac{x-1}{5}\right)^2+\left(\frac{y+4}{5}\right)^2=1\)
\(\Rightarrow\) Đặt \(\left\{{}\begin{matrix}\frac{x-1}{5}=sina\\\frac{y+4}{5}=cosa\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5sina+1\\y=5cosa-4\end{matrix}\right.\) thế vào B:
\(B=\left(5sina-6\right)^2+\left(5cosa-8\right)^2\)
\(B=25sin^2a+25cos^2a-60sina-80cosa+100\)
\(B=125-100\left(\frac{3}{5}sina+\frac{4}{5}cosa\right)=125-100.sin\left(a+\alpha\right)\)
\(\Rightarrow B_{min}=25\) khi \(sin\left(a+\alpha\right)=1\) \(\Rightarrow\left\{{}\begin{matrix}sina=\frac{3}{5}\\cosa=\frac{4}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5sina+1=4\\y=5cosa-4=0\end{matrix}\right.\) \(\Rightarrow N\left(4;0\right)\)
