JEEometry

${ y }^{ 2 }=4ax\\ y=mx-2am-a{ m }^{ 3 }\quad (equation\quad of\quad normal\quad with\quad slope\quad 'm')\quad \\ a=1\\ \Rightarrow y=mx-2m-{ m }^{ 3 }\quad \\ P(h,k).\\ \Rightarrow { m }^{ 3 }+(2-x)m-k=0,\quad roots\quad are\quad { m }_{ 1 },{ m }_{ 2 },{ m }_{ 3 }.\\ \Rightarrow k={ m }_{ 1 }{ m }_{ 2 }{ m }_{ 3 }=\alpha { m }_{ 3 }......(1)\\ \Rightarrow (2-x)={ m }_{ 1 }{ m }_{ 2 }+{ m }_{ 1 }{ m }_{ 3 }+{ m }_{ 2 }{ m }_{ 3 }=\alpha -{ { m }_{ 3 } }^{ 2 }\quad (\because { m }_{ 1 }+{ m }_{ 2 }=-{ m }_{ 3 })\\ \Rightarrow { { m }_{ 3 } }^{ 2 }=\alpha -2+x\quad \quad ......(2)\\ From\quad (1)\& (2):\\ { k }^{ 2 }={ \alpha }^{ 2 }(\alpha -2+x)\\ But\quad this\quad is\quad part\quad of\quad { y }^{ 2 }=4ax\\ \Rightarrow \alpha =\boxed { 2 } \\$