\(c^2=a^2+b^2=16+9=25\Rightarrow c=5\)
\(\Rightarrow F\left(5;0\right)\)
Do đường thẳng \(\Delta\) qua F và vuông góc Ox nên có pt: \(x-5=0\)
Tọa độ A; B là nghiệm: \(\left\{{}\begin{matrix}\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{25}{16}-\dfrac{y^2}{9}=1\\x=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y^2=\dfrac{81}{16}\\x=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\pm\dfrac{9}{4}\\x=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}A\left(5;-\dfrac{9}{4}\right)\\B\left(5;\dfrac{9}{4}\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{AB}=\left(0;\dfrac{9}{2}\right)\Rightarrow AB=\sqrt{0^2+\left(\dfrac{9}{2}\right)^2}=\dfrac{9}{2}\)