\(\left\{{}\begin{matrix}\overrightarrow{PA}=\left(a-4;-3\right)\\\overrightarrow{PB}=\left(-4;b-3\right)\end{matrix}\right.\)
\(PA\perp PB\Rightarrow\overrightarrow{PA}.\overrightarrow{PB}=0\)
\(\Rightarrow-4\left(a-4\right)-3\left(b-3\right)=0\)
\(\Rightarrow\left(b-3\right)=-\dfrac{4}{3}\left(a-4\right)\)
Đồng thời \(b-3=-\dfrac{4}{3}\left(a-4\right)\Rightarrow b=-\dfrac{4}{3}\left(a-4\right)+3=-\dfrac{4}{3}a+\dfrac{25}{3}\)
Mà \(b\ge0\Rightarrow-\dfrac{4}{3}a+\dfrac{25}{3}\ge0\Rightarrow a\le\dfrac{25}{4}\)
\(S_{PAB}=\dfrac{1}{2}PA.PB=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+9}.\sqrt{16+\left(b-3\right)^2}\)
\(=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+9}.\sqrt{\dfrac{16}{9}\left(a-4\right)^2+16}\)
\(=2\sqrt{\left[\left(a-4\right)^2+9\right]^2}=2\left(a-4\right)^2+18\)
\(\Rightarrow\dfrac{S}{2}=f\left(a\right)=a^2-8a+25\)
Xét hàm \(f\left(a\right)=a^2-8a+25\) trên \(\left[0;\dfrac{25}{4}\right]\)
\(-\dfrac{b}{2a}=4\) ; \(f\left(0\right)=25\) ; \(f\left(4\right)=9\) ; \(f\left(\dfrac{25}{4}\right)=\dfrac{225}{16}\)
\(\Rightarrow f\left(a\right)_{max}=f\left(0\right)=25\)
\(\Rightarrow a=0\Rightarrow b=\dfrac{25}{3}\Rightarrow T=2.0+3.\dfrac{25}{3}=25\)
