It's intuitively obvious to me that only two tangents can be drawn to a parabola from a point $(a,b)$ outside it but I want a mathematical pre-calculus proof of this.
Let the parabola be given by $y=ax^2+bx+c$, and let the exterior point be $(x_0,y_0)$. A line through that point is $y=m(x-x_0)+y_0$. To look for intersections, we solve the parabola equation and the line equation simultaneously:
$$ax^2+bx+c = m(x-x_0)+y_0$$
$$ax^2 + (b-m)x+(c+mx_0-y_0)=0$$
If the line is tangent to the parabola, then this equation has exactly one solution, which means its discriminant must equal $0$. The discriminant is:
$$B^2-4AC = (b-m)^2-4a(c+mx_0-y_0)$$
Setting this equal to $0$, and taking $m$ as our variable, we get:
$$m^2 - (2b + 4ax_0)m + (b^2 - 4ac + 4ay_0)=0$$
Being quadratic, this equation can have, at most, two solutions for $m$.
Does that work for you?
As a side note, that final quadratic could also have one solution, or no solutions. These situations correspond to the cases where $(x_0,y_0)$ is actually on the parabola, or respectively, inside the parabola.
Say $\ell :ax+by+c=0$ is tangent to (given) parabola $y^2=2px$. We are seeking for $a,b,c$, not all $0$, given $p\ne 0$. We can assume $a=1$. Then the quadratic equation
$$y^2+2pby+2pc =0$$ must have only one solution on $y$, so $D=0$ and we get
$$ 4p^2b^2-8pc =0$$ so $pb^2 = 2c\;\;\;(*)$. Since the point $T(x_0,y_0)$ is on tangent we have $c=-x_0-by_0$. Pluging in to $(*)$ we have: $$ pb^2 = -2x_0-2by_0$$
So this equation is quadratic on $b$ so it has at most two solution and thus we have at most 2 tangents (no matter where $T$ lies).