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Does anyone know how to find this? I know I have enough information to uniquely generate an ellipse (you only need five points for a conic, I have three plus the slopes or tangents at those points which count as points), but my problem is trying to solve the system of equations that include the derivatives that blow up to infinity because one of the tangents is vertical.

Here are my points and the tangents associated with them.

\begin{align} P_1&=(0,-3);\ \ m_1=+1\\ P_2&=(1,\pm0);\ \ m_2=\infty \; \; \text{i.e. $(x=1)$}\\ P_3&=(0,+3);\ \ m_3=-1\\ \end{align}

Thank you for helping to solve this problem!

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    $\begingroup$ Instead of derivatives ($dy/dx$), you can use differentials ($dx$ and $dy$) thusly: If this is the equation of the conic ... $$A x^2 + 2 B x y + C y^2 + 2 D x + 2 E y + F = 0$$ ... then this is the differential ... $$2 A x\;dx + 2 B ( x\;dy + y\;dx ) + 2 C y\;dy + 2 D\;dx + 2 E\;dy = 0$$ Now, you can make appropriate substitutions for the slopes without fear of blowing-up: $$\begin{align}m_1:&\quad dx = 1, dy = 1 \\ m_2:&\quad dx = 0, dy = 1 \\ m_3:&\quad dx = 1, dy = -1\end{align}$$ (That said, the conic is determined using the 3 pts & only 2 tangents. So, you can just ignore the vertical.) $\endgroup$ – Blue Jan 26 '18 at 6:02
  • $\begingroup$ @Blue However, it’s worth checking once you’ve computed the ellipse that the tangent at $P_2$ is vertical. $\endgroup$ – amd Jan 26 '18 at 22:23
  • $\begingroup$ @amd: Yes, checking the third tangent (whichever one the "third" might be) is a good idea. I didn't say so in the comment (ran out of space), but you may notice that I did so in my determinant-based answer. $\endgroup$ – Blue Jan 26 '18 at 23:17
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A computationally-expensive approach (that avoids solving a linear system of six unknowns with only five variables) is to consider that the equation of the conic through the points $P=(P_x, P_y)$, $Q=(Q_x,Q_y)$, $R=(R_x,R_y)$, $S=(S_x,S_y)$, $T=(T_x,T_y)$ is given by

$$\left|\begin{array}{c,c,c,c,c,c} x^2 & y^2 & x y & x & y & 1 \\ P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\ Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\ R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\ S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\ T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\ \end{array}\right| = 0$$

You are given three points, say, $P = (0,-3)$, $Q = (1,0)$, $R=(0,3)$. You need two more. We can approximate them using small displacements along any two of the tangent lines. For instance, $$m_1 = 1: \;S = P + s\,(1,1) = (s,s-3) \qquad m_2 = \infty: T = Q + t\,(0,1) = (1,t)$$

Expanding the determinant with the help of a tool such as Mathematica, yields

$$-6 s t \left(\quad\begin{array}{c} (3+s+3t-st) x^2 + t(s-1) x y - (s-1) y^2 \\ + ( 6-10s-3t+st) x + 9 (s-1) \end{array}\quad\right)= 0$$

Using typical Calculus chicanery, we insist that $s$ and $t$ are merely small, not zero, so that we can divide them out of the equation ...

$$(3+s+3t-st) x^2 + t(s-1) x y - (s-1) y^2 + ( 6-10s-3t+st) x + 9 (s-1)= 0$$

... and then proceed to calculate the limiting form of the equation as $s$ and $t$ become vanishingly small ... by substituting $s=t=0$!

$$3 x^2 + y^2 + 6 x - 9 = 0$$

Note that, upon differentiation, we have $$6 x + 2 y y^\prime + 6 = 0$$ which is satisfied by $(x,y) = (0,3)$ and $y^\prime = -1$.

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Start with something relatively easy—find the circle that passes through $P_1$ and $P_3$ and has the appropriate tangents. Its center is the intersection of the perpendiculars to the tangents through the two points. By symmetry, that’s at $(-3,0)$, and its radius is then $3\sqrt2$. Let $$f: (x,y)\mapsto(x+3)^2+y^2-18$$ so that the equation of this circle is $f(x,y)=0.$ Now let $$g:(x,y)\mapsto(x+y-3)(x-y-3).$$ The degenerate conic $g(x,y)=0$ consists of the tangent lines through $P_1$ and $P_2$ (it’s the product of the equations of the lines). Every nontrivial linear combination of these two equations is itself a conic that passes through the two points and has the correct tangents at those points. Use Plücker’s mu to find the linear combination that also passes through $P_2$: $$f(P_2)g(x,y)-g(P_2)f(x,y) = -2(x+y-3)(x-y-3)-4((x+3)^2+y^2-18)=0$$ which simplifies to the equation $$3x^2+y^2+6x-9=0.$$ Happily, this is an ellipse. The gradient at $(1,0)$ is $(12,0)$, so the tangent at $P_2$ is vertical, as required. (As Blue noted in his comment, we didn’t need to use all three tangents to construct this ellipse.)

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