Parabola

[man111]

The latus rectum of the parabola whose parametric coordiantes are [tex]x=at^2+bt+c[/tex] and [tex]y=a't^2+b't+c'[/tex]

[nevrodermit]

If we had the equation of the parabola in the form [tex](x-h)^2 = 4p(y - k)[/tex] then the length of the latus rectum is [tex]|4p|[/tex].

So now we have to convert the given parametric equations to the one above.
From [tex](x-h)^2 = 4p(y - k)[/tex] we have [tex]x^2-2xh+h^2=4py-4pk\Leftrightarrow x^2-2xh+h^2+4pk=4py \Leftrightarrow y=\frac{1}{4p}x^2-\frac{1}{2p}x+\frac{h^2+4pk}{4p}[/tex]

Let [tex]y=mx^2+nx+l[/tex] and thus
[tex]\begin{array}{|l} m= \frac{1}{4p} \\ n=-\frac{1}{2p} \\ l=\frac{h^2+4pk}{4p} \end{array}[/tex]
We need to express the coefficients [tex]m, n, l[/tex] with [tex]a,b,c,a', b', c'[/tex].

We are given
[tex]\begin{array}{|l} x=at^2+bt+c \\ y=a't^2+b't+c' \end{array}[/tex]
So [tex]a't^2+b't+c'=m(at^2+bt+c)^2+n(at^2+bt+c)+l[/tex] which is true for all values of [tex]t[/tex]. Expanding it we get
[tex]a't^2+b't+c'=m(a^2t^4+b^2t^2+c^2+2abt^3+2bct+2act^2)+nat^2+nbt+nc+l[/tex]
[tex]a't^2+b't+c'=ma^2t^4+mb^2t^2+mc^2+2mabt^3+2mbct+2mact^2+nat^2+nbt+nc+l[/tex]
[tex]a't^2+b't+c'=ma^2t^4 + 2mabt^3+t^2(mb^2+2mac+na)+t(2mbc+nb)+mc^2+nc+l[/tex]
So we get
[tex]\begin{array}{|l} ma^2=0 \\2mab = 0 \\ a'= mb^2+2mac+na \\ b'= 2mbc+nb \\ c'=mc^2+nc+l \end{array}[/tex]

Clearly [tex]m\ne 0 \Rightarrow a=0[/tex]. The system becomes
[tex]\begin{array}{|l} a'= mb^2 \\ b'= 2mbc+nb \\ c'=mc^2+nc+l \end{array}[/tex]

[tex]m=\frac{a'}{b^2}[/tex], [tex]4p=\frac{1}{m}=\frac{b^2}{a'}[/tex] so the answer is [tex]|4p|=b^2|(a')^{-1}|[/tex]