Straight Line Touches Parabola

The condition for a line y = mx + c to be 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 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 equal real roots, then the line (ii) will intersect the parabola (i) at one point only and so 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 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.



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