Lets say that we can draw a tangent line to parabola from point #P(a,b)#. Then, there exists point #Q(r,s)# in parabola #f(x)=x^2# which our tangent line goes through. Because #Qinf(x)#, it follows that:
#f(r) = r^2 => s=r^2 => Q(r,r^2)#
The equation of the tangent line is: #y=kx+n#, where #k# is slope and:
#k=f'(x)=2x#
Because our tangent line goes through #Q#, it follows that: #k=2r# and if we insert coordinates of #Q# and #k# into the equation of the tangent line, we get:
#y=kx+n#
#r^2=2r*r+n => n = -r^2#
So, our tangent line is: #y=2rx-r^2#.
We started with assumptation that #P in y# (tangent line exists and goes through #P# and #Q#). Then (insert coordinates of #P# into the equation of tangent line),
#b=2ra-r^2 => r^2-2ra+b=0#
The last equation gives us the answer: is it possible to construct the tangent line from arbitrary point #P# to parabola #f(x)=x^2#? If so, there exists the solution of the quadratic equation #r^2-2ra+b=0#. Because this is a quadratic equation, we may get one solution, two solutions or no solution based on the coefficients of the equation.
So, let solve it:
#r^2-2ra+b=0 <=>#
#<=> r^2-2ra + a^2 - a^2+b=0 <=>#
#<=> (r-a)^2 - (sqrt(a^2-b))^2 = 0 <=>#
#<=> (r-a-sqrt(a^2-b))(r-a+sqrt(a^2-b))=0 <=>#
#<=> r-a-sqrt(a^2-b) = 0 vv r-a+sqrt(a^2-b)=0 <=>#
#<=> r=a+sqrt(a^2-b) vv r=a-sqrt(a^2-b)#
Real solutions exist when #a^2-b>=0#, or #b<=a^2#.
When #a^2=b#, we get one solution:
#r=a#.
When #a^2>b#, we get two solutions:
# r=a+sqrt(a^2-b) #
#r=a-sqrt(a^2-b)#
Finally, when #b>a^2# we don't have solutions.
