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The problem asks to find find equations for the two lines through the point $(3, 13)$ that are tangent to the parabola $y=6x-x^2$.

I'm trying to play with finding slopes and points of tangency but then I asked myself if there are only two tangents through that point or if there are infinitely many and the problem asks to find two of them. I do understand that the in the problem description suggests there are only two but I don't know how to prove it mathematically (if that's true).

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  • $\begingroup$ Hint: Start by finding the equation for a line tangent to the parabola at a point $(x_{0},6x_{0}-x_{0}^{2})$ on the parabola, and use that to find which lines also go through $(3,13)$. $\endgroup$ – DMcMor Jun 28 '17 at 18:57
  • $\begingroup$ Or simply notice that the abscissa of the vertex of the parabola is $3$, so the wanted tangents are straightforward to find through Archimedes' lemma - they are symmetric with respect to the axis of the parabola. $\endgroup$ – Jack D'Aurizio Jun 28 '17 at 18:58
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Let us call the point on the parabola $(x,y)$. Since it must be on a tangent line to the parabola, we find the following equations:

$$y = -x^2 + 6x \tag{1}$$

$$\frac{y-13}{x-3} = \frac{d(-x^2 + 6x)}{dx} = -2x + 6 \iff y = -2x^2 + 12x - 5 \tag{2}$$

From (1) and (2), it follows that:

$$-x^2 + 6x = -2x^2 + 12x - 5 \iff x^2 - 6x + 5 = 0 \iff x = 3 \pm 2$$

We thus find two solutions: $(1,5)$ and $(5,5)$.

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The geometric construction of the tangents to a parabola from a point $P$ that lies on the axis of the parabola is fairly easy:

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If $P'$ is the symmetric of $P$ with respect to the vertex $V$ of the parabola and $A,B$ are the points on the parabola such that $P'\in AB$ and $AB\perp PP'$, the tangents through $P$ are exactly given by $PA$ and $PB$. This (Archimedes' Lemma) is the geometric counter-part of $\frac{d}{dx}\,x^2 = 2x$.

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You will have some equations for coefficients of this line. Let it be $y = kx +b$. Let $x_0$ is point of contact. So $k \cdot x_0 + b = 6 x_0 - x_0^2; k = -2x_0$ ($x_0 = -k/2$; $-k^2/2 + b = -3k -k^2/4$). And so we have equations $13 = 3k + b$ and $-k^2/2 + b = -3k -k^2/4$. Solutions are $k = -2\cdot\sqrt{13}, b = 6\cdot\sqrt{13} + 13$ and $k = 2\cdot\sqrt{13}, b = -6\cdot\sqrt{13} + 13$.

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Axiomatically, through any two points, there is exactly one line. By the definition of tangency, a line that is tangent to a curve touches that curve at exactly one point. Because the given point outside of the curve lies exactly at the axis of symmetry of the parabola, through that point, and the point of tangency on one side of the parabola, exactly one line will exist. Because of symmetry, the same will happen at the other side. Any other point that would be a part of any other line would have to be collinear with any of the previous two, therefore, there are only those two tangents through the given point.

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    $\begingroup$ Your definition of tangency works for parabolas, but not generally. E.g., $y=x^3$ would have a tangent line only at the origin with this definition. $\endgroup$ – amd Jun 28 '17 at 22:34
  • $\begingroup$ I absolutely agree! I offered this response specifically addressing the given problem, as the question references the problem description as inspiration for wanting a proof. Providing a general proof would be ideal, but I'm not sure I am skilled enough to make it intuitive, immediate, relevant, and useful to the person who asked. Thank you for the very correct comment! $\endgroup$ – Axiomaric Jun 28 '17 at 22:37

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