elimination-of-ordinate

Elimination of the Ordinate

Summary: Closed-form eliminants for the ordinate $y$ between two curve equations of low degree in $y$, derived by repeated subtraction of suitably-multiplied copies of the two equations. Euler tabulates the cases (deg $y$ = 1 vs $n$), (2,2), (2,3), (3,3), (4,4); the last is built by a cascade of brevity substitutions $A,B,C,D\mid a,b,c,d\mid E,F,G\mid H,I,h,i$ that exploits the recursive symmetries $a=D$, $G=e$, $I=h$.

Sources: chapter19 (§§474–482).

Last updated: 2026-05-12.

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Throughout, $P,Q,R,S,T,\dots$ and $p,q,r,s,t,\dots$ denote non-irrational (polynomial or rational) functions of $z$ alone — i.e., everything except the powers of $y$. The pair of curves is written

$\text{I: }P+Qy+R{y}^2+S{y}^3+T{y}^4+\cdots =0,\text{II: }p+qy+r{y}^2+s{y}^3+t{y}^4+\cdots =0.$

Degree (1, $n$) in $y$ (§§474–477)

If one curve has $y$ only to first power, $P+Qy=0$, then $y=-P/Q$. Substituting into the second equation eliminates $y$ outright.

• (1, 1). $P+Qy=0$, $p+qy=0⟹pQ-Pq=0$.

• (1, 2). $P+Qy=0$, $p+qy+r{y}^2=0⟹P^{2}r-PQq-p{Q}^2=0$, i.e.,

  $P^{2}r-PQq+Q^{2}p=0\text{ (after sign rearrangement)}.$

  Euler’s derivation goes via the intermediate step III: $(Pq-pQ)+Pry=0$, then multiply by $Pr$ and combine with the first.

• (1, 3). Substituting $y=-P/Q$ into the cubic gives

  $Q^{3}p-P{Q}^{2}q+P^{2}Qr-P^{3}s=0.$

• (1, 4).

  $Q^{4}p-P{Q}^{3}q+P^2{Q}^{2}r-P^{3}Qs+P^{4}t=0.$

In every case, $y=-P/Q$ is a single-valued function of $z$, so by the §470 criterion every real root of the eliminant corresponds to a real intersection.

Degree (2, 2) in $y$ (§§478–479)

Pure quadratics $P+R{y}^2=0$, $p+r{y}^2=0$: subtracting suitably gives

$Pr-Rp=0.$

A real root $z_0$ of this corresponds to two intersections, $y=\pm \sqrt{-P/R}$, provided $-P/R>0$ at $z_0$; if $-P/R<0$, the intersections are complex (§478). Two roots above and below the axis are forced only when the axis is a diameter of both curves.

General quadratics $P+Qy+R{y}^2=0$, $p+qy+r{y}^2=0$: the elimination cascades through

$\text{III: }(Pq-Qp)+(Pr-Rp)y=0,\text{IV: }(Pr-Rp)+(Qr-Rq)y=0,$

and equating $y$ from both gives

$\boxed{P^2{r}^2-2PRpr+R^2{p}^2+Q^{2}pr-PQqr-QRpq+PR{q}^2=0.}$

Each real root corresponds to a real $y$ from III or IV — unless III and IV share a polynomial factor, in which case division could reveal hidden complex intersections (see complex-intersections §473).

Degree (2, 3) in $y$ (§480)

Both curves I: $P+Qy+R{y}^2=0$ and II: $p+qy+r{y}^2+s{y}^3=0$. Reduce II to degree 2 using I:

$\text{III: }(Pq-Qp)+(Pr-Rp)y+Ps{y}^2=0.$

Then apply the (2,2) recipe to I and III to obtain a single equation in $z$ alone. The result is a polynomial in $P,Q,R,p,q,r,s$ of fifteen-plus terms whose explicit form is recorded in §480 (after dividing through by $P$):

$P^3{s}^2-2P^{2}Rqs-P^{2}Qrs+3PQRps+P{Q}^{2}qs-Q^{3}ps+R^3{p}^3$ $+P^{3}R{r}^2-PQRqr-2P{R}^{2}pr+Q^{2}Rpr+P{R}^2{q}^2-Q{R}^{2}pq=0.$

(Euler’s printed form contains the same fifteen terms; sign or coefficient discrepancies should be checked against the source if used.)

Degree (3, 3) in $y$ (§481)

Both cubics, $P+Qy+R{y}^2+S{y}^3=0$ and $p+qy+r{y}^2+s{y}^3=0$. Multiply I by $p$, II by $P$, subtract and divide by $y$:

$\text{III: }(Pq-Qp)+(Pr-Rp)y+(Ps-Sp)y^2=0.$

Multiply I by $s$, II by $S$, subtract:

$\text{IV: }(Sp-Ps)+(Sq-Qs)y+(Sr-Rs)y^2=0.$

Compare to the (2,2) general-quadratic case under the identification

$P\to Pq-Qp,Q\to Pr-Rp,R\to Ps-Sp,p\to Sp-Ps,q\to Sq-Qs,r\to Sr-Rs.$

The resulting expression contains seven terms, six of which are divisible by $Sp-Ps$; the seventh combines with the unmatched term to give a divisor $Sp-Ps$ overall, allowing a long division. The final expression (printed in full in §481) runs to over thirty terms in $P,Q,R,S,p,q,r,s$.

Degree (4, 4) in $y$ — cascade of brevity substitutions (§482)

Both quartics. Euler treats this as a masterclass in cascaded elimination. The procedure has four passes, each reducing the degree by one via brevity letters that hide the symmetric polynomial structure.

Pass 1. Multiply I by $p$, II by $P$, subtract and divide by $y$:

$\text{III: }(Pq-Qp)+(Pr-Rp)y+(Ps-Sp)y^2+(Pt-Tp)y^3=0.$

Multiply I by $t$, II by $T$, subtract:

$\text{IV: }(Pt-Tp)+(Qt-Tq)y+(Rt-Tr)y^2+(St-Ts)y^3=0.$

Name $A=Pq-Qp$, $B=Pr-Rp$, $C=Ps-Sp$, $D=Pt-Tp$ for III, and $a=Pt-Tp$, $b=Qt-Tq$, $c=Rt-Tr$, $d=St-Ts$ for IV. Note: $a=D$, and Euler verifies

$Ad-Cb=(Pt-Tp)(Ps-Qs)=D\alpha ,Ac-Bb=(Pt-Tp)(Rq-Qr)=D\beta ,$

with $\alpha =Sq-Qs$, $\beta =Rq-Qr$ — additional identifications that close the recursion.

Pass 2. III and IV are now both cubics in $y$:

${\text{III}}^{\prime }:A+By+C{y}^2+D{y}^3=0,{\text{IV}}^{\prime }:a+by+c{y}^2+d{y}^3=0.$

Multiply III’ by $d$, IV’ by $D$, subtract and divide:

$\text{V: }(Ad-Da)+(Bd-Db)y+(Cd-Dc)y^2=0.$

Multiply III’ by $a$, IV’ by $A$, subtract:

$\text{VI: }(Ab-Ba)+(Ac-Ca)y+(Ad-Da)y^2=0.$

Name $E=Ab-Ba$, $F=Ac-Ca$, $G=Ad-Da$ for VI, and $e=Ad-Da$, $f=Bd-Db$, $g=Cd-Dc$ for V. Then $G=e$, and again Euler records $Eg-Ff=G\zeta$ with $\zeta =Cb-Bc$, so $Eg-Ff$ is divisible by $G$.

Pass 3. V and VI are quadratics in $y$:

${\text{V}}^{\prime }:e+fy+g{y}^2=0,{\text{VI}}^{\prime }:E+Fy+G{y}^2=0.$

Apply the (1,2)–style elimination:

$\text{VII: }(Ef-Fe)+(Eg-Ge)y=0,\text{VIII: }(Eg-Ge)+(Fg-Gf)y=0.$

Name $H=Ef-Fe$, $I=Eg-Ge$ for VII and $h=Eg-Ge$, $i=Fg-Gf$ for VIII. Then $I=h$, giving

$\boxed{Hi-Ih=0}$

— an equation in $z$ alone, after unrolling all substitutions. Each unrolling absorbs one layer of $A$–$D$, $a$–$d$, $E$–$G$, $e$–$g$, and finally the original $P,Q,R,S,T,p,q,r,s,t$. Euler notes that the equation in $E,F,G,e,f,g$ is divisible by $G=e$, and the equation in $A,B,C,D,a,b,c,d$ is divisible by $D^2=a^2$, so the final equation reduces to one in eight letters — four upper-case and four lower-case (§482 closing).

The general $(m,n)$ case (§482 closing)

In general, by this method, whatever the power of $y$, both equations may contain, the variable $y$ can always be eliminated to find an equation which contains only $z$. (source: chapter19, §482)

The cascade $(\text{deg }m,\text{deg }n)\to (\text{deg }m-1,\text{deg }n-1)\to \cdots$ terminates in a linear pair, whose elimination is immediate. Each step doubles the size of the symbolic expression, motivating the alternative method by indeterminate multipliers of §§483–485, which avoids the repeated substitutions.

Related pages

• intersection-of-two-curves
• complex-intersections
• indeterminate-multiplier-elimination
• chapter-19-on-the-intersection-of-curves