The latus rectum of the parabola defined parametrically by $x=at^2+bt+c$ and $y=a't^2+b't+c'$ is---

I tried to eliminate $t$ from both the equations and I got the following equation $$(a'x-ay+c'a-ca')^2=(ab'-ba')(b'x-by+c'b-cb')$$

However I could not find out the length of latus rectum. I also tried to find out the equation of axis using this but it turned out a bit complex. Please help me in this regard. Thanks. :)

  • $\begingroup$ If you just want the length of the latus rectum (had to look this term up), you could try taking your $t$ elimination and rotating the coordinate system. By geometric considerations the axis of the parabola will have slope $a'/a$, so if you apply a rotation to $x$ and $y$, to make the axis vertical, you might get a simpler expression of type $Y = AX^2 + BX + C$, from which I assume you can find the latus easily. $\endgroup$ – Erick Wong Dec 2 '16 at 18:38
  • $\begingroup$ There's probably a more elegant geometric way to do it, but this seems like it will concretely work. $\endgroup$ – Erick Wong Dec 2 '16 at 18:38
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    $\begingroup$ @ErickWong One can also find the direction of the axis of symmetry by finding the kernel of the matrix of the quadratic part $\pmatrix{a^2&-aa'\\-aa&a'^2}$. $\endgroup$ – amd Dec 2 '16 at 18:41
  • $\begingroup$ Oops. I have the diagonal elements transposed in that matrix. $\endgroup$ – amd Dec 2 '16 at 18:56
  • $\begingroup$ ... and left out a prime in the lower-left corner. That’ll teach me to comment on the run. $\endgroup$ – amd Dec 3 '16 at 7:19

You’ve done some nice work in finding the Cartesian equation for the parabola, but it think it might be easier overall to work with the parametric form $\mathbf r:t\mapsto(x(t),y(t))$. (I’ve omitted most of the tedious details of the algebraic manipulations in the following.)

The first order of business is to find the angle of the parabola’s axis of symmetry. One way to do this is to use the fact that for any two points on a parabola, the line defined by their midpoint and the intersection of the tangents at the two points is parallel to this axis. Taking $t=\pm1$ is reasonably convenient. The midpoint is simply $(\mathbf r(1)+\mathbf r(-1))/2$. For the intersection of the tangents, we have $\mathbf r(1)+s\mathbf r'(1)=\mathbf r(-1)+t\mathbf r'(-1)$, which expands into the system $$\begin{align}s(2a+b)+a+b+c &= t(-2a+b)+a-b+c \\ s(2a'+b')+a'+b'+c'&=t(-2a'+b')+a'-b'+c\end{align}$$ with solution $s=-1$, $t=1$. The direction vector for the parabola’s axis is thus $$\begin{align}\frac12(\mathbf r(1)+\mathbf r(-1))-(\mathbf r(-1)+\mathbf r'(-1)) &= \frac12(\mathbf r(1)-\mathbf r(-1))-\mathbf r'(-1) \\ &=(b,b')-(b-2a,b'-2a') \\ &=(2a,2a'),\end{align}$$ so we can take $\mathbf n=(a,a')$ as the direction of the axis.

Next, we find the parabola’s vertex. The tangent to the parabola at the vertex is orthogonal to its axis, which gives rise to the equation $$\mathbf n\cdot\mathbf r'(t)=(a,a')\cdot(at+b,a't+b')=a(at+b)+a'(a't+b')=0.$$ Solving for $t$ we get $$t_c=-\frac12{ab+a'b'\over a^2+a'^2}\tag{1}.$$ You can verify that the function $\mathbf r$ is symmetric with respect to this point in the sense that the chord defined by the points $\mathbf r(t_c\pm\Delta t)$ is orthogonal to the parabola’s axis, so the two points are equidistant from the vertex.

The tangents at the ends of the latus rectum meet the axis at a 45° angle, which means that they are orthogonal to each other. This fact leads to the equation $$\mathbf r'(t_c+\Delta t)\cdot\mathbf r'(t_c-\Delta t)=(2a(t_c+\Delta t)+b)(2a(t_c-\Delta t)+b)+(2a'(t_c+\Delta t)+b')(2a'(t_c-\Delta t)+b')=0$$ for the ends of the latus rectum, which has the solutions $$\Delta t=\pm\frac12{ab'-a'b\over a^2+a'^2}\tag{2}.$$ Because of the symmetry noted previously, we know that both of these solutions represent the same pair of points.

Finally, we compute the length of the latus rectum: $$\begin{align}\|\mathbf r(t_c+\Delta t)-\mathbf r(t_c-\Delta t)\|^2 &= (\mathbf r(t_c+\Delta t)-\mathbf r(t_c-\Delta t))\cdot(\mathbf r(t_c+\Delta t)-\mathbf r(t_c-\Delta t)) \\ &= \left({a'(ab'-a'b)^2\over(a^2+a'^2)^2}\right)^2+\left({a(ab'-a'b)^2\over(a^2+a'^2)^2}\right)^2 \\ &={(ab'-a'b)^4\over(a^2+a'^2)^3},\end{align}$$ so the length of the latus rectum is $${(ab'-a'b)^2\over(a^2+a'^2)^{3/2}}.\tag{3}$$ With this distance in hand, you can now easily find the parabola’s focus and directrix, if needed.

Incidentally, this is another path to a Cartesian equation for this parabola. If you have the directrix given by an equation in the form $\mathbf n\cdot\mathbf x=d$ and the focus $\mathbf f$, then an equation of the parabola is $$\left({d-\mathbf n\cdot\mathbf x\over\|\mathbf n\|}\right)^2=(\mathbf x-\mathbf f)\cdot(\mathbf x-\mathbf f).$$

To continue working with the Cartesian equation instead, I’d take a slightly different approach than in the cited paper. First look at the equation of a parabola with axis parallel to the $y$-axis, $y=ax^2+bx+c$. For such a parabola, the length of the latus rectum is simply $|1/a|$. For the general parabola $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, we take Erick Wong’s suggestion to rotate so as to eliminate the quadratic terms involving $y$. The equation will then be of the form $A'x'^2+D'x'+E'y'+F=0$ (note that the constant term is unchanged by a rotation), with latus rectum length $|E'/A'|$. Invariance of the trace tells us that $A'=A+C$, but finding $E'$ will take a a bit more work.

For a parabola, $B^2=4AC$, so we can rewrite the general equation as $(\alpha x+\beta y)^2+Dx+Ey+F=0$, first multiplying through by $-1$ if necessary to make $A$ and $C$ positive. In matrix form this is $\begin{bmatrix}x&y&1\end{bmatrix}M\begin{bmatrix}x&y&1\end{bmatrix}^T=0$, with $$M=\begin{bmatrix}\alpha^2&\alpha\beta&D/2\\\alpha\beta&\beta^2&E/2\\D/2&E/2&F\end{bmatrix}.$$ The direction of the parabola’s axis is given by an eigenvector of $0$ (i.e., an element of the kernel) of the quadratic part of this matrix, one of which is $\begin{bmatrix}-\beta&\alpha\end{bmatrix}^T$. We want to rotate so as to bring this vector parallel to the $y$-axis. The appropriate rotation is $$R=\begin{bmatrix}{\alpha\over\sqrt{\alpha^2+\beta^2}}&-{\beta\over\sqrt{\alpha^2+\beta^2}}&0\\{\beta\over\sqrt{\alpha^2+\beta^2}}&{\alpha\over\sqrt{\alpha^2+\beta^2}}&0\\0&0&1\end{bmatrix}$$ and $$R^TMR = \begin{bmatrix}\alpha^2+\beta^2&0&\frac12{\alpha D+\beta E\over\sqrt{\alpha^2+\beta^2}} \\ 0&0&\frac12{\alpha E-\beta D\over\sqrt{\alpha^2+\beta^2}} \\ \frac12{\alpha D+\beta E\over\sqrt{\alpha^2+\beta^2}} & \frac12{\alpha E-\beta D\over\sqrt{\alpha^2+\beta^2}} & F\end{bmatrix}.$$ Comparing this to the result in the first paragraph above, we find that the latus rectum length for the general equation is $${|\alpha E-\beta D|\over(\alpha^2+\beta^2)^{3/2}} = {|E\sqrt A\mp D\sqrt C|\over(A+C)^{3/2}}.$$ Choose the sign opposite to that of $B$. Many of the terms in the Cartesian equation you derived are constant, so pulling the coefficients of $x$ and $y$ out of it to construct $D$ and $E$ doesn’t look too bad.

  • $\begingroup$ Nice approach. However, the appearance of $\bullet ^{3/2}$ in high school/college maths/physics is quite rare (the only example I can think of is Kepler's Third Law). Could you please take a look at your equation $(3)$ vs my solution and point out if there is any discrepancy in mine? Thanks. $\endgroup$ – hypergeometric Dec 9 '16 at 16:21
  • $\begingroup$ @hypergeometric Off the top off my head, I’d say that you’re starting off with a huge assumption: that the $t$’s on both sides of the initial equivalence are the same. There’s no reason for that to be true. The r.h.s. is one specific parameterization of a parabola, but as you’ve seen from the Bézier curves, there are in fact an infinite number of them that all take the form of a pair of quadratics. Calling it the parametrization might be leading you astray. $\endgroup$ – amd Dec 9 '16 at 19:08
  • $\begingroup$ Notwithstanding the earlier comment on $3/2$, I think your answer is correct. However, even if they the huge assumption is not always true, it does not mean it is always untrue. Hence the equivalence should hold in one specific case. $\endgroup$ – hypergeometric Dec 10 '16 at 0:14
  • $\begingroup$ @hypergeometric The exponent in the denominator seems plausible. Once you’ve rotated and translated the parabola into standard position, the l.r. length is the ratio of the coefficients of $y$ and $x^2$. The latter is just $a^2+a'^2$, while the rotation introduces a factor of $sqrt{a^2+a'^2}$ when the direction vector is normalized to get the sine and cosine of parabola’s axis angle. $\endgroup$ – amd Dec 10 '16 at 3:44
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    $\begingroup$ An interesting feature of expression (3) is that it’s invariant with respect to affine transformations of the parameter. That is, if you replace $t$ by $ht'+k$ $(h\ne0)$ and plug the resulting coefficients into (3), $h$ and $k$ fall out of the expression upon simplification, leaving (3) again. All such parameterizations of a parabola are related by affine transformations; see this question. $\endgroup$ – amd Dec 11 '16 at 17:26

Equating the parametric form of the parabola given to the parametric form for a general parabola in the solution here (and putting $M=AE-CD$): $$\bigg(at^2+bt+c, \;\;a't^2+b't+c\bigg)\equiv\left(\frac{Ct^2-Et+CF}M,\;\; -\frac {At^2-Dt+AF}M\right)$$ Equating coefficients of $t$ for the $x, y$ components respectively: $$\begin{align} a=\;\;\frac CM, &\qquad a'=-\frac AM\\ b=-\frac EM, &\qquad b'=\;\;\frac DM\\ c=\;\;\frac{CF}M &\qquad c'=-\frac {AF}M \end{align}$$ Note that $$\hspace {2cm}a'b-ab'=\frac {AE-CD}{M^2}=\frac 1M\qquad\cdots (1)$$ and $$\hspace {2cm}a'^2+a^2=\frac {A^2+C^2}{M^2}=\frac K{\;M^2}\qquad\cdots (2)$$ where $K=A^2+C^2$.

From the solution here the length of the Latus Rectum of a general parabola $(Ax+Cy)^2+Dx+Ey+F=0$ is given as: $$\begin{align} \text { Length of Latus Rectum }&=\frac {\big|AE-CD\big|}{\;\;\big(A^2+C^2\big)^\frac 32}\\ &=\frac M{\;\;K^\frac 32}\\ &=\frac{\left(\displaystyle\frac 1M\right)^2}{\;\left(\displaystyle\frac K{\;M^2}\right)^\frac 32}\qquad\qquad (1)^2\div(2)^\frac 32\\ &=\color{red}{\frac{\big(a'b-ab'\big)^2}{\big(a'^2+a^2\big)^\frac 32}}\end{align}$$


Following from the recent solution here on the general form of a parametric parabola we can provide a more direct derivation.

Use the notation $p=b', q=a'$.

Let $\lambda=pb-qa$ and $\mu=pc-ar$.

Note that the general form of the parametric parabola is $(Ax+Cy)^2+Dx+Ey+F+0$ where

$$\begin{align} A&=\;\;p,\\ C&=-a,\\ E&=-(b\lambda-2a\mu),\\ D&=\;\;\;q\lambda-2p\mu\end{align}$$

From the solution here,

$$\begin{align} \text{Length of Latus Rectum }&= \frac {|AE-CD|}{(A^2+C^2)^{\frac 32}}\\ &=\frac{\big|-p\big[b\lambda-2a\mu\big]-(-a)\big[q\lambda-2p\mu\big]\big|}{\big(p^2+(-a)^2\big)^{\frac 32}}\\ &=\frac{\big|-(pb-aq)\lambda\big|}{(p^2+a^2)^{\frac 32}}\\ &=\frac {(pb-aq)^2}{(p^2+a^2)^{\frac 32}}\\ &=\color{red}{\frac {(a'b-ab')^2}{(a'^2+a^2)^{\frac 32}}} \end{align}$$

  • $\begingroup$ This is a solution for only one specific parameterization, given on the r.h.s. of your first equivalence. There are an infinite number of them, however, so without any other information there’s no good reason to believe that the parameter $t$ on both sides of that equivalence is the same. $\endgroup$ – amd Dec 9 '16 at 19:10
  • $\begingroup$ @amd - Does that mean there is no one specific way to specify the length of the latus rectum for a given parameterization and the answer is indeterminate? Even if so, in the particular case that they happen to be the same, it would appear that the answer is applicable. $\endgroup$ – hypergeometric Dec 9 '16 at 23:58
  • $\begingroup$ That’s not at all what I’m saying. I’ve given you a general solution for any parameterization as a pair of quadratics. What you’ve done here (modulo using the questionable result for the general equation) is rewritten the solution for one specific parameterization by renaming some of the quantities involved. $\endgroup$ – amd Dec 10 '16 at 3:38
  • $\begingroup$ That said, at least as far as the Bézier parameterizations go, I’m pretty sure that they can all be converted into this one via an affine transformation of the parameter. However, I’m not as sure that this exhausts all of the quadratic parameterizations. $\endgroup$ – amd Dec 10 '16 at 3:39
  • $\begingroup$ Your formula for the l.r. length gives the wrong answer for the basic parabola $y=ax^2$ (see my comment there), so that makes the conclusion dubious as well. $\endgroup$ – amd Dec 10 '16 at 3:41

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