# Bài 48 trang 114 SGK Hình học 10 Nâng cao

##### Hướng dẫn giải

a) Giả sử: $$M\left( {x;y} \right) \in \left( C \right)$$

\eqalign{ & MF = \sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {1 - y} \right)}^2}}\cr&d\left( {M,\Delta } \right) = {{|x + y - 1|} \over {\sqrt 2 }} \cr & {{MF} \over {d\left( {M,\Delta } \right)}} = e = 1\cr& \Leftrightarrow \sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {1 - y} \right)}^2}} = {{|x + y - 1|} \over {\sqrt 2 }} \cr & \Leftrightarrow 2\left( {{x^2} - 2x + 1 + {y^2} - 2y + 1} \right) = \cr&\;\;\;\;\;\;\;\;\;{x^2} + {y^2} + 1 + 2xy - 2x - 2y \cr & \Leftrightarrow {x^2} + {y^2} - 2xy - 2x - 2y + 3 = 0 \cr}

\eqalign{ & b)\,\,\,{{MF} \over {d\left( {M,\Delta } \right)}} = \sqrt 2 \cr&\Leftrightarrow \sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {1 - y} \right)}^2}} = |x + y - 1| \cr & \Leftrightarrow {x^2} - 2x + 1 + {y^2} - 2y + 1 = \cr&\;\;\;\;{x^2} + {y^2} + 1 + 2xy - 2x - 2y \cr & \Leftrightarrow 2xy - 1 = 0 \cr}

\eqalign{ & c)\,\,\,{{MF} \over {d\left( {M,\Delta } \right)}} = {1 \over {\sqrt 2 }}\cr& \Leftrightarrow \sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {1 - y} \right)}^2}} = {{|x + y - 1|} \over 2} \cr & \Leftrightarrow 4\left( {{x^2} - 2x + 1 + {y^2} - 2y + 1} \right) = \cr&\;\;\;\;\;{x^2} + {y^2} + 1 + 2xy - 2x - 2y \cr & \Leftrightarrow 3{x^2} + 3{y^2} - 6x - 6y - 2xy + 7 = 0. \cr}