$\text{ The straight line }y=m(x-a)\text{ meets the parabola }{y}^2=4ax\text{ in }$$\text{ two distinct points for }=$

$\begin{array}{l}\text{substitute the expression for }y\text{\hspace{0.17em} in the equation of the parabola}\\ {\left(m\left(x-a\right)\right)}^2=4ax\\ m^2{\left(x-a\right)}^2=4ax\\ m^2{x}^2-4ax-2a{m}^{2}x+a^2{m}^2=0\\ m^2{x}^2+x\left(-4a-2a{m}^2\right)+a^2{m}^2=0\\ \text{This is a quadratic equation, if it has two roots, its discriminant must be positive}\\ {\left(4a+2a{m}^2\right)}^2-4\left(m^2\right)\left(a^2{m}^2\right)>0\\ {\left(2+m^2\right)}^2-m^4>0\\ m^4+4m^2+4-m^4>0\\ 4\left(m^2+1\right)>0\end{array}$

$For any non zero value of m this is possible$