Normal Parabola

$We\quad take\quad a\quad point\quad { P }_{ 1 }=({ x }_{ 0 },{ x }_{ 0 }^{ 2 })\quad on\quad the\quad parabola.\\ Then\quad slope\quad of\quad tangent\quad is=2{ x }_{ 0 }\quad Hence\quad slope\quad \\ of\quad normal\quad is\quad \frac { -1 }{ 2{ x }_{ 0 } } .\quad So\quad equation\quad of\quad normal\\ is\quad (x-{ x }_{ 0 })=-2{ x }_{ 0 }(y-{ x }_{ 0 }^{ 2 }).\quad Solving\quad it\quad with\quad the\quad parabola\\ we\quad get\quad x={ x }_{ 0 },-\frac { 1 }{ 2{ x }_{ 0 } } -{ x }_{ 0 }\quad .\quad So\quad the\quad other\quad point\quad is\\ { P }_{ 2 }=(-\frac { 1 }{ 2{ x }_{ 0 } } -{ x }_{ 0 },{ (-\frac { 1 }{ 2{ x }_{ 0 } } -{ x }_{ 0 } })^{ 2 })\\ So\quad { P }_{ 1 }{ P }_{ 2 }=\sqrt { 4{ x }_{ 0 }^{ 2 }{ (1+\frac { 1 }{ 4{ x }_{ 0 }^{ 2 } } ) }^{ 3 } } .\quad For\quad maximising\quad { P }_{ 1 }{ P }_{ 2 }\\ we\quad will\quad maximise\quad the\quad part\quad under\quad the\quad radical.\\ f\left( { x }_{ 0 } \right) =4{ x }_{ 0 }^{ 2 }{ (1+\frac { 1 }{ 4{ x }_{ 0 }^{ 2 } } ) }^{ 3 }.\quad Equating\quad it's\quad derivative\quad to\quad 0\\ we\quad get\quad { x }_{ 0 }=\pm \frac { 1 }{ \sqrt { 2 } } .Put\quad the\quad value\quad and\quad we\quad get\quad \\ { P }_{ 1 }{ P }_{ 2 }=\frac { 3\sqrt { 3 } }{ 2 } .\\ \\$