# Straight Line Touches a Parabola

The condition for a line $y = mx + c$ to be a tangent to the parabola ${y^2} = 4ax$ is that $c = \frac{a}{m}$ and the tangent to the parabola is $y = mx + \frac{a}{m}$.

Consider that the standard equation of a parabola with vertex at origin $\left( {0,0} \right)$ 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 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 we 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. If equation (iii) has equal real roots, then the line (ii) will intersect the parabola (i) at one point only and thus is the tangent to the parabola..

Equation (iii) will have equal roots if

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

This is the condition for the line (ii) to be a tangent to the parabola (i). Putting this value of $c$ in equation (ii), we have
$y = mx + \frac{a}{m}$

This is the tangent to the parabola.