# Intersection of a Line and a Parabola

The line $$y = mx + c$$ intersects the parabola $${y^2} = 4ax$$ at two points maximum and the condition for such intersection is that $$a > mc$$.

Consider the standard equation of a parabola with the vertex at origin $$\left( {0,0} \right)$$, which can be written as

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

Also the equation of a line is represented by

\[y = mx + c\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)\]

To find the point of the intersection of the parabola (i) and the given line (ii), using the method of solving simultaneous equations we solve equation (i) and equation (ii), in which one equation is in quadratic form and the other is in linear form. So take the value of $$y$$ from equation (ii) and put this value in equation (i), i.e. the equation of the 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 $$x$$, we can solve this quadratic equation either by completing the square method or by using the quadratic formula, and can have at most two roots (i.e. values of $$x$$), and we put the values of $$x$$ in equation (ii) to get the values of $$y$$ and obtained two points. This shows that the line (ii) can intersect the parabola (i) at two points maximum. It is also clear from the given diagram.

Equation (iii) will have two real roots if:

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