Bài 4 trang 67

a) Ta có:

\(\begin{array}{l}AF = \sqrt {{{\left( { - 4 + 3} \right)}^2} + {{\left( {0 - 1} \right)}^2}}  = \sqrt 2 ,d\left( {A,\Delta } \right) = \frac{{\left| { - 3 + 0.1 + 5} \right|}}{{\sqrt {{1^2} + {0^2}} }} = 2 \Rightarrow \frac{{AF}}{{d\left( {A,\Delta } \right)}} = \frac{{\sqrt 2 }}{2}\\BF = \sqrt {{{\left( { - 4 - 2} \right)}^2} + {{\left( {0 - 8} \right)}^2}}  = 10,d\left( {B,\Delta } \right) = \frac{{\left| {2 + 0.8 + 5} \right|}}{{\sqrt {{1^2} + {0^2}} }} = 7 \Rightarrow \frac{{BF}}{{d\left( {B,\Delta } \right)}} = \frac{{10}}{7}\\CF = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {0 - 3} \right)}^2}}  = 5,d\left( {C,\Delta } \right) = \frac{{\left| {0 + 0.3 + 5} \right|}}{{\sqrt {{1^2} + {0^2}} }} = 5 \Rightarrow \frac{{CF}}{{d\left( {C,\Delta } \right)}} = 1\end{array}\)

b)

+ Vì \(\frac{{AF}}{{d\left( {A,\Delta } \right)}} < 1\) nên A nằm trên elip

+ Vì \(\frac{{BF}}{{d\left( {B,\Delta } \right)}} > 1\) nên B nằm trên hypebol

+ Vì \(\frac{{CF}}{{d\left( {C,\Delta } \right)}} = 1\) nên C nằm trên parabol