Respuesta :

Answer:

[tex]{ \tt{p(x) : {y}^{2} = 4ax}} \\ { \tt{f(x) : {(y')}^{2} = 4a {x}'}} \\ { \tt{g(x) : {(y'')}^{2} = 4ax''}} \\ { \bf{since \: they've \: a \: common \: root}} : \\ {y}^{ 2} = {(y')}^{2} = {(y'')}^{2} \\ = > { \tt{4ax + 4ax' + 4ax''}} \\ = 4a(x + x' + x'') \\ common \: root \: is \: 4a \\ { \tt{ \infin {}^{ - } \leqslant 4a \leqslant \infin {}^{ + } }}[/tex]

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