Question 1024736
{{{x^2=4ay}}}, or a better form with which to work, {{{y=(1/(4a))x^2}}}.  The line or lines tangent to the parabola will have slope {{{dy/dx=(1/(4a))*2x=(1/(2a))x}}}.


Any general point on the parabola is  (x, (x^2)/(4a)).  
Pick any of these points you like as long as  {{{x<>0}}}.  Why?  Because at x=0, the slope is 0, for a horizontal line, and will not make an angle with the x-axis.  The question description specifies that the tangent line makes an angle with the x-axis.