I get totally stuck in this exercise:

Consider the set of parabolas with vertex in $V=(1,1)$ and, as axis, the line $y=2x-1$. Write the equation of the generic parabola of that set.

Thank you for your help!

  • $\begingroup$ Where do you get stuck? What have you tried so far? $\endgroup$ – Taroccoesbrocco May 15 '18 at 9:45
  • $\begingroup$ @Taroccoesbrocco I know how to find the equation of a parabola given vertex and focus, or directrix, or a point on the patabola. But with vertex and axis I have no idea. $\endgroup$ – F.inc May 15 '18 at 9:56
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    $\begingroup$ It is a SET of parabolas: that means, for instance, that you can choose a focus on the axis at will. $\endgroup$ – Intelligenti pauca May 15 '18 at 10:24
  • $\begingroup$ @Aretino got that. But, after having chosen a focus is it possible to write the equation of a generic parabola and not that, of that one? $\endgroup$ – F.inc May 15 '18 at 11:32
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    $\begingroup$ It is possible, if you parameterize the position of the focus (e.g. $F=(t, 2t-1)$) and write the equation of the parabola for a generic value of $t$. $\endgroup$ – Intelligenti pauca May 15 '18 at 11:48

There are many ways to attack this. Two in particular allow you to pretty much write an equation down by inspection.

Affine transformation: A generic equation of a parabola with vertex at the origin and the $x$-axis as its axis of symmetry is $y^2=ax$. We seek an affine transformation $A$ that will map the $x$-axis onto the line $y=2x-1$ and the origin onto $(1,1)$. In addition, $A$ must map the $y$-axis onto the perpendicular line $x+2y=3$ to preserve the correct tangent at the vertex. Since affine transformations preserve incidence, this will also take care of mapping the origin correctly. If $C$ is the matrix of a conic, it transforms as $A^{-T}CA^{-1}$, so we’re really more interested in finding the inverse transformation.

Fortunately, lines are covariant—they also transform via $A^{-T}$, so we can construct this matrix directly: its first column is the image of the $y$-axis, its second is the image of the $x$-axis and its third the image of the line at infinity, which is mapped to itself by any affine transformation. Therefore, $$A^{-1} = \begin{bmatrix}1&2&-3\\2&-1&-1\\0&0&1\end{bmatrix}.$$ Instead of computing $A^{-T}CA^{-1}$ and then converting that into an equation, it’s much simpler to just make the appropriate substitutions into the generic equation from the preceding paragraph, producing the transformed equation $$(2x-y-1)^2=a(x+2y-3).\tag1$$ Rearrange as needed. The only actual computation that was required here was determining the constant term of the tangent line equation.

If you prefer, you can start instead from the canonical $y^2=4px$. If you normalize the equations of the two lines, $A$ will then be an isometry, preserving the geometric meaning of the parameter $p$.

Focus-directrix: Choose a convenient parameterization of the axis line, such as $f:\lambda\mapsto(1+\lambda,1+2\lambda)$, which is symmetric about the vertex $(1,1)$. If $f(\lambda)$ is the focus of a parabola with that vertex, then its directrix is the line $x+2y=3-5\lambda$. A parabola is the locus of points equidistant from the focus and directrix, so using well-known formulas for these distances and squaring them, we can immediately write the equation $$\frac15(x+2y+5\lambda-3)^2 = (x-1-\lambda)^2+(y-1-2\lambda)^2$$ for this parabola. Again, we only needed to work out the constant term of the directrix equation and normalize it, after which we could write down an equation for the parabola directly. Collecting all of the terms involving $\lambda$ on one side and factoring gives $$\frac15(2x-y-1)^2=4\lambda(x+2y-3)$$ which only differs from equation (1) by a constant factor.

Lubin mentions in a comment trying to attack this projectively by using the fact that parabolas are tangent to the line at infinity. This approach requires a bit more work than the previous two, but isn’t too bad.

Polarities: A parabola is tangent to the line at infinity where its axis intersects that line. This pole-polar pair together with the known tangent at the vertex is enough to find the coefficients of the equations of the family of parabolas. If $C$ is the matrix of the parabola, each pole-polar pair $(\mathbf p,\mathbf l)$ satisfies the equation $\mathbf l\times C\mathbf p=0$, which generates three linear equations in the elements of $C$, only two of which are independent. Finding the coefficients of the parabola’s equation then reduces to solving a system of linear equations.

Since $C$ is symmetric, it only has six unique elements. Let $$C=\begin{bmatrix}c_{11}&c_{12}&c_{13}\\c_{12}&c_{22}&c_{23}\\c_{13}&c_{23}&c_{33}\end{bmatrix}.$$ The intersection of the parabola’s axis with the line at infinity is just its direction vector, so we have $$\begin{bmatrix}0\\0\\1\end{bmatrix}\times C\begin{bmatrix}1\\2\\0\end{bmatrix} = \begin{bmatrix} -c_{12}-2c_{22} \\ c_{11}+2c_{12} \\ 0 \end{bmatrix}$$ and $$\begin{bmatrix}1\\2\\-3\end{bmatrix}\times C\begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix} 3c_{12}+2c_{13}+3c_{22}+5c_{23}+2c_{33} \\ -3c_{11}-3c_{12}-4c_{13}-c_{23}-c_{33} \\ -2c_{11}-c_{12}-2_{c13}+c_{22}+c_{23} \end{bmatrix}.$$ The solution set of the resulting homogeneous system is then the null space of the matrix $$\begin{bmatrix} 0 & -1 & 0 & -2 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 3 & 2 & 3 & 5 & 2 \\ -3 & -3 & -4 & 0 & -1 & -1 \\ -2 & -1 & -2 & 1 & 1 & 0 \end{bmatrix}.$$ We know that the last three rows are not independent, so any one of them can be omitted to simplify the computation. The null space is spanned by $\mathbf v_1 = (8,-4,-5,2,0,8)^T$ and $\mathbf v_2 = (24,-12,-11,6,8,0)^T$, but since we’re working with homogeneous coordinates, we need only consider affine combinations $(1-\mu)\mathbf v_1+\mu\mathbf v_2$. After removing a constant common factor, an equation of the family of parabolas is therefore $$(1-\mu)(4x^2-4xy+y^2-5x+4)+\mu(12x^2-12xy+3y^2-11x+8y)=0,$$ which I don’t find nearly as transparent as the equations derived using the preceding methods. On the other hand, it includes the degenerate conics that represent both the axis and vertex tangent lines (for $\mu=1/2$ and $\mu=-1/2$, respectively), which the other families don’t.

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  • $\begingroup$ One question: in the third way (polarities), why the direction vector of the axis $y=2x-1$ is $[1, 2, 0]$ and not $[-2, 1, 1]$, did you choose a normal one? Thank you for your exhaustive comment! $\endgroup$ – F.inc May 21 '18 at 8:35
  • $\begingroup$ @F.inc $[-2:1:1]$ is the homogeneous representation of the line, and directly corresponds to the equation $-2x+y+1=0$, or $(-2,1)\cdot(x,y)=-1$. $(-2,1)$ is normal to this line and the line’s direction vectors are orthogonal to this one. Moreover, we need a point on the line for the constraint, not a line. This line’s point at infinity is $[-2:1:1]\times[0:0:1]=[1:2:0]$, which also represents the line’s direction. $\endgroup$ – amd May 21 '18 at 18:06
  • $\begingroup$ @F.inc As a point of technique, if there’s a possibility of my getting points and lines confused when working with homogeneous coordinates, I’ll use column vectors for points and row vectors for lines. Then if I accidentally use the wrong type of object, the matrix products won’t make sense. $\endgroup$ – amd May 21 '18 at 18:14

Consider a parabola $y-1=A(x-1)^2$ rotated clockwise by $\alpha$ about the point $(1,1)$.

The equation of the rotated parabola is $$v=Au^2$$ where $$\begin{align} \left(\begin{array} . u\\v\end{array}\right) &= \left(\begin{array}.\;\ \cos(-\alpha)&\sin(-\alpha)\\-\sin(-\alpha)&\cos(-\alpha)\end{array}\right) \left(x-1\atop y-1\right)\\ &=\left(\begin{array}. \cos\alpha&-\sin\alpha\\\sin\alpha&\;\;\ \cos\alpha\end{array}\right) \left(x-1\atop y-1\right)\\ &=\left(\begin{array}. \sin\theta&-\cos\theta\\\cos\theta&\;\;\ \sin\theta\end{array}\right) \left(x-1\atop y-1\right) &&\scriptsize(\alpha+\theta=\tfrac\pi 2)\\ \end{align}$$ and $\tan\theta=2$ (slope of $y=2x-1$).

Hence equation of rotated parabola is

$$\begin{align} (x-1)\cos\theta+(y-1)\sin\theta &=A\big[(x-1)\sin\theta-(y-1)\cos\theta\big]^2\\ \tfrac 1{\sqrt{5}}(x-1)+\tfrac 2{\sqrt{5}}(y-1) &=A\big[\tfrac 2{\sqrt{5}}(x-1)-\tfrac 1{\sqrt{5}}(y-1)\big]^2\\ \tfrac 1{\sqrt{5}}\big[(x-1)+2(y-1)\big]&=\tfrac 15 A\big[2(x-1)-(y-1)\big]^2\\ x+2y-3 &=\tfrac 1{\sqrt{5}}A\big[2x-y-1\big]^2\\ \color{red}{x+2y-3} &\color{red}{=A'\big[2x-y-1\big]^2}\\ \end{align}$$ where $A, A'$ are constants.

See Desmos implementation here.

enter image description here

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    $\begingroup$ This is good. I tried to attack it projectively, using the fact that parabolas are characterized as the conics that are tangent to the line at infinity, but I got bogged down in computation when I tried to specify the axis. +1. $\endgroup$ – Lubin May 16 '18 at 18:54
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    $\begingroup$ @hypergeometric thank you for that brilliant answer $\endgroup$ – F.inc May 17 '18 at 17:16
  • $\begingroup$ @F.inc - Thanks. Glad you liked it :) $\endgroup$ – Hypergeometricx May 17 '18 at 17:17

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