Latus rectum of a parametric parabola

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.

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: $$\begin{align} \text { Length of Latus Rectum }&=\frac {\big|AE-CD\big|}{\;\;\big(A^2+C^2\big)^\frac 32}\\ &=\frac M{\;\;K^\frac 32}\\ &=\frac{\left(\displaystyle\frac 1M\right)^2}{\;\left(\displaystyle\frac K{\;M^2}\right)^\frac 32}\qquad\qquad (1)^2\div(2)^\frac 32\\ &=\color{red}{\frac{\big(a'b-ab'\big)^2}{\big(a'^2+a^2\big)^\frac 32}}\end{align}$$

Addendum

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,

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