Intersection of Line and Parabola

The line y = mx + c intersects the parabola {y^2} = 4ax at most two points and condition for such intersection is that a  > mc.
Consider the standard equation of parabola with vertex at origin \left(  {0,0} \right)can be written as

{y^2}  = 4ax\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)

Also equation of a line is represented by

y =  mx + c\,\,\,{\text{ -  -  - }}\left( {{\text{ii}}} \right)

To find the point of intersection of parabola (i) and the given line (ii), using the method of solving simultaneous equation we solve equation (i) and equation (ii), in which one equation is in quadratic and other is in linear form, so take value of y from equation (ii) and putting this value in equation (i) i.e. equation of parabola becomes


\begin{gathered} {\left( {mx + c} \right)^2} = 4ax \\ \Rightarrow {m^2}{x^2} + 2mcx + {c^2} = 4ax \\ \Rightarrow {m^2}{x^2} + 2\left( {mc - 4a}  \right)x + {c^2} = 0\,\,\,{\text{ -   -  - }}\left( {{\text{iii}}}  \right) \\ \end{gathered}

Since equation (iii) is a quadratic equation in x, and we can solve this quadratic equation either by completing square method or using quadratic formula and can have at most two roots i.e. values of x and putting values of xin equation (ii) to get the values of y and obtained two points. This shows that the line (ii) can intersect the parabola (i) at most two points. It is also clear from the given diagram.
Equation (iii) will have two real roots if

\begin{gathered} {\text{Discriminant  >   0}} \\ \Rightarrow {\left( {2mc - 4a} \right)^2} -  4{m^2}{c^2} > 0 \\ \Rightarrow 4{m^2}{c^2} - 16mca + 16{a^2} -  4{m^2}{c^2} > 0 \\ \Rightarrow   - 16mca + 16{a^2} > 0 \\ \Rightarrow 16{a^2} > 16mca \\ \Rightarrow \boxed{a > mc} \\ \end{gathered}



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