No Intersection between Line and Parabola

The line y = mx + c does not intersects the parabola {y^2} = 4ax if 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. If equation (iii) has imaginary roots, then the line (ii) will not intersect the parabola (i), it is clear from the given diagram.
Equation (iii) will have imaginary 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}

Which is the required condition line is not intersects the parabola.



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