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I'd like to find the equation of the parabola that passes through two points, and for which the slope of the parabola is defined at those two points; I believe that these conditions fully define, but do not over-define the parabola. Note that, in general, the axis of symmetry of the parabola won't be parallel to the $x$ or $y$ axis, so that the equation can't be written as $ax^2 + bx + c - y = 0$. As an example, consider the parabola that passes through point $A$ at $(-95,3)$ and is tangent (at $A$) to a line with slope of $35^{\circ}$ to the x axis. It also passes through $B$ at $(-150,50)$ and is tangent (at $B$) to a line with slope of $8^{\circ}$ to the x axis. What's the equation of the parabola that satisfies these conditions? I believe that the general equation of a parabola is $ax^2 + bxy + cy^2 + dx + ey + f = 0$. I also understand that I need to differentiate at the tangency points to find two linear equations, but I'm unsure how to find my six unknowns ($a$ through $f$) from the conditions that define my parabola. Many thanks in advance!

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  • $\begingroup$ Solving the system can be problematic, since you don't know which coefficient(s) might vanish. A "direct" (though computationally-expensive) approach, described in this answer, uses a $6\times 6$ determinant for the five-point conic. Two points are your givens ($A$ and $B$); two more are infinitesimally-displaced from those along the tangent lines. You'll need a fifth point, which geometry provides: If $M$ is midpoint of $A$ and $B$, and $N$ is the point where the tangent lines meet, then the midpoint of $M$ and $N$ lies on the parabola. $\endgroup$ – Blue Sep 13 '19 at 23:49
  • $\begingroup$ Effectively a duplicate of math.stackexchange.com/q/3206543/265466. See also math.stackexchange.com/q/2622143/265466 and math.stackexchange.com/q/2999165/265466. $\endgroup$ – amd Sep 14 '19 at 4:05
  • $\begingroup$ Note that you can reduce the number of unknowns immediately. Since you know that the curve is a parabola, the quadratic part must be a perfect square, so you can replace it with $(\alpha x+\beta y)^2$. $\endgroup$ – amd Sep 14 '19 at 6:17
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Expanding upon my comment, and following the same basic strategy as in this answer, we can get the equation mechanically using the determinant for the five-point conic through $P=(P_x,P_y)$, $Q=(Q_x,Q_y)$, $R=(R_x,R_y)$, $S=(S_x,S_y)$, $T=(T_x,T_y)$: $$\left|\begin{array}{c,c,c,c,c,c} x^2 & y^2 & x y & x & y & 1 \\ P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\ Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\ R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\ S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\ T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\ \end{array}\right| = 0 \tag{$\star$}$$

It happens to be convenient to have the origin at the midpoint of the given points, so let us take $$P = (d\cos\theta,d\sin\theta) \qquad Q = (-d\cos\theta, -d\sin\theta) \tag{1}$$ Two more points come from infinitesimally-displacing $P$ and $Q$ along their tangent lines (in directions, say, $\phi$ and $\psi$, respectively): $$R = P + (p\cos\phi, p\sin\phi) \qquad S = Q + (q\cos\psi,q\sin\psi) \tag{2}$$ for "very small" $p$ and $q$. The fifth point is provided by the specific geometry of the parabola; if $M$ is the midpoint of $P$ and $Q$, and $N$ is the point where the tangent lines at $P$ and $Q$ meet, then the midpoint of $M$ and $N$ lies on the parabola. Thus, we can take $$T = \frac{d}{2\sin(\phi-\psi)}\left(\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta), \sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta)\right) \tag{3}$$ Substituting into $(\star)$, and letting a computer algebra system crunch some symbols, we factor-out and cancel a $p$ and $q$, then set the remaining $p$s and $q$s to zero. Canceling factors of $d^2 \csc^2(\phi-\psi) \sin(\phi-\theta)\sin(\psi - \theta)$, we have

$$\begin{align} 0 &= \phantom{2}x^2 (\sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta))^2 \\ &+ \phantom{2}y^2 (\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta))^2 \\ &-2x y (\sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta))(\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta)) \\ &-4 x d \sin\theta \sin(\phi-\psi) \sin(\phi-\theta) \sin(\psi-\theta) \\ &+4 y d \cos\theta \sin(\phi-\psi) \sin(\phi-\theta) \sin(\psi-\theta) \\ &-4 d^2 \sin^2(\phi-\theta)\sin^2(\psi-\theta) \end{align} \tag{$\star\star$}$$

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  • $\begingroup$ Thanks Blue, that's brilliant. I really appreciate you working so hard on this. So applying this solution to my original example, I would "move" the origin to the mid-point between A and B, which I find to be at (-122.5,26.5). That means that P = (27.5,-23.5) and Q = (-27.5, 23.5). I then find that d = 36.1732 and theta = -0.7071 radians. I originally defined A as being at 35° to the x axis (I realized I didn't specify whether it was a positive or negative slope, but let's assume negative for both angles), so phi = -0.6109 radians and psi = -0.1396 radians. Does this seem correct so far? $\endgroup$ – Nexus Sep 14 '19 at 3:39
  • $\begingroup$ As a check, I use the x co-ordinates of points P & Q in the beautiful equation you found, but I don't find that the six terms sum to zero. This is likely an error on my part, but I wanted to ask if (i) I followed your solution correctly up to the point of solving the equation and (ii) whether you found all terms to sum to zero using P and Q as test cases. Many thanks again, Blue - you're a champion. $\endgroup$ – Nexus Sep 14 '19 at 3:42
  • $\begingroup$ @Nexus: I agree with all of your calculations. Inputting them into $(\star\star)$ with Mathematica, I get that the parabola's equation is $$217.074+36.927x-1.38651x^2+43.2124y-4.72149xy-4.01955y^2=0$$ Entering the $x$ and $y$ coordinates of $P$ yields $-4.54747\times 10^{-13}$, which, given the numerical approximations involved, counts as zero. ... I've double-checked that I didn't make a typo. ... You can tell that $(x,y)=d(\cos\theta,\sin\theta)$ "should" work: the linear terms instantly cancel; the second-degree terms factor as a perfect square that simplifies to cancel with the constant. $\endgroup$ – Blue Sep 14 '19 at 4:36
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    $\begingroup$ and one more for the constant term: -4*(36.1732)^2*(sin(-0.6109--0.7071))^2*(sin(-0.1396--0.7071))^2 = -13.9701 (but you get 217.074). I won't write out the other three coefficients, for which I also get different answers - I suspect those should be enough to trouble-shoot what's going wrong. Thanks again, Blue - I really appreciate you sticking with this problem! $\endgroup$ – Nexus Sep 14 '19 at 15:24
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    $\begingroup$ Brilliant - that's helped me zero in on my error, which was a typo in the xy coefficient (which I thought I'd thoroughly checked...). Thanks so much, Blue - I really appreciate your help, and all the hard work. $\endgroup$ – Nexus Sep 14 '19 at 16:45

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