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.

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    $\begingroup$ Nothing wrong with the geometric proof. Draw a line through the chosen point which doesn't intersect the parabola. Rotate the line on the point until it "hits" (touches) the parabola. Rotate it the other way until it "hits" again. Those are the two tangents. (We didn't prove that they will be tangents, but we proved that no other lines through the chosen point can be tangents.) $\endgroup$ – Wildcard Oct 11 '17 at 23:47

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).

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    $\begingroup$ this is by no means a solution to the problem : you don't adress the question. You have to start from an exterior point $(x_0,y_0)$ and then show, as @G Tony Jacobs has done, that there are two tangents issued from this point ! I am very surprized that 5 people have upvoted this answer ! Unless you modify it, I will downvote it. $\endgroup$ – Jean Marie Oct 11 '17 at 22:27
  • $\begingroup$ To expand on @JeanMarie's complaint, you have showed that there can only be two tangent lines described with a given fixed $a$ and $c$ (which I guess do determine the lines' $x$-intercept), not two tangent lines through an arbitrary given point. $\endgroup$ – aschepler Oct 12 '17 at 0:09

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