Subtangent and Subnormal

Summary: From the linear truncation $At+Bu=0$ at the point $M=(p,q)$, the subtangent $PT=-Bq/A$ is the segment on the axis from $P$ (foot of ordinate at $M$) to $T$ (where the tangent meets the axis). Dropping the normal at $M$ to the axis (rectangular coordinates) gives the subnormal $PN=-Aq/B$, with $PN=P{M}^2/PT$. Tangent length $MT=\sqrt{P{T}^2+P{M}^2}$ and normal length $MN=(PM/PT)\sqrt{P{T}^2+P{M}^2}$. Three worked examples: the parabola yields $PT=2p$ (recovering [[parabola|chapter 6’s $PT=2x$]]); the ellipse yields $PT=-(a^2-p^2)/p$; a third-order seventh-species cubic yields $PT=2p^2{q}^2/(a{p}^2-c)$.

Sources: chapter13 (§§289, 292). Figures 55, 59, 60 (in figures55-60).

Last updated: 2026-05-07.

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Subtangent (§289)

The tangent at $M$ meets the original axis at a point $T$. From the similar right triangles $Mq\mu$ (with legs $Mq=t$, $q\mu =u$) and $MPT$ (with legs $PM=q$, $PT$):

$\frac{q\mu }{Mq}=\frac{MP}{PT}⟹\frac{u}{t}=\frac{q}{PT}.$

But $u/t=-A/B$ from the linear truncation, so

$\boxed{PT=-\frac{Bq}{A}.}$

The portion $PT$ of the original axis between $P$ and $T$ is called the subtangent. The sign records on which side of $P$ the point $T$ lies.

Rule for finding the subtangent (§289)

In the equation for the curve, after finding that the ordinate $y=q$ corresponds to the abscissa $x=p$, substitute $x=p+t$ and $y=q+u$. After the substitution, keep only the terms in which $t$ and $u$ appear with exponent 1, and neglect all other terms, to obtain the equation $At+Bu=0$. From this we know $A$ and $B$, and so the subtangent $PT=-Bq/A$.

Worked examples (§289)

Example I — parabola

Curve: $y^2=2ax$ (vertex at origin, principal axis as $x$-axis).

Take $AP=p$, $PM=q$, so $q^2=2ap$. Substitute $x=p+t$, $y=q+u$: $q^2+2qu+u^2=2ap+2at.$

Keep only the linear terms: $2qu-2at=0$, so $u/t=a/q$, giving $A=-a$, $B=q$ (up to a common factor 2). Then

$PT=-\frac{B}{A}q=\frac{q^2}{a}=\frac{2ap}{a}=2p.$

So the subtangent of the parabola is twice the abscissa — exactly [[parabola|the §150 result $PT=2x$ from chapter 6]].

Example II — ellipse

Curve: $a^2{y}^2+b^2{x}^2=a^2{b}^2$ (center at origin, principal axes).

Take $AP=p$, $PM=q$, so $a^2{q}^2+b^2{p}^2=a^2{b}^2$. Substitute $x=p+t$, $y=q+u$ and keep linear terms: $2a^{2}qu+2b^{2}pt=0,$

so $u/t=-b^{2}p/(a^{2}q)$, giving $A=b^{2}p$, $B=a^{2}q$. Then

$PT=-\frac{B}{A}q=-\frac{a^2{q}^2}{b^{2}p}=-\frac{a^2-p^2}{p}$

(using $a^2{q}^2=b^2(a^2-p^2)$). The negative sign places $T$ on the opposite side of $P$ from $A$, in agreement with the ellipse tangent geometry of chapter 6.

Example III — third-order line of the seventh species

Curve: $y^{2}x=a{x}^2+bx+c$.

Take $AP=p$, $PM=q$, so $p{q}^2=a{p}^2+bp+c$. Substitute $x=p+t$, $y=q+u$ and expand: $(p+t)(q^2+2qu+u^2)=a(p^2+2pt+t^2)+b(p+t)+c.$

Drop superfluous terms to obtain $2pqu+q^{2}t=2apt+bt$, i.e.,

$\frac{u}{t}=\frac{2ap+b-q^2}{2pq},A=-(2ap+b-q^2),B=2pq.$

Then

$PT=-\frac{B}{A}q=\frac{2p{q}^2}{2ap+b-q^2}=\frac{2p^2{q}^2}{a{p}^2-c}$

(after multiplying top and bottom by $p$ and using $p{q}^2=a{p}^2+bp+c$ to write $2ap+b-q^2=(a{p}^2-c)/p$ via $bp=p{q}^2-a{p}^2-c$, so $2ap+b-q^2=(a{p}^2-c)/p$).

Subnormal (§292)

If the original coordinates are rectangular, the line $MN$ through $M$ perpendicular to the tangent meets the axis at $N$ — this is the normal. From similar right triangles $Mq\mu$ (legs $t,u$) and $MPN$ (legs $q$, $PN$):

$\frac{Mq}{q\mu }=\frac{MP}{PN}⟹\frac{t}{u}=\frac{q}{PN},$

so

$\boxed{PN=-\frac{Aq}{B}.}$

The portion $PN$ between the foot of ordinate $P$ and the foot of normal $N$ is the subnormal.

From subtangent (§292)

In rectangular coordinates the subnormal can be obtained from the subtangent without redoing the similar-triangle argument. Right triangles $TPM$ and $MPN$ share the altitude $PM$, so

$\frac{PT}{PM}=\frac{PM}{PN}⟹PN=\frac{P{M}^2}{PT}.$

This is the geometric-mean relation between the two segments cut off by the foot of the altitude in a right triangle.

Tangent and normal lengths (§292)

If the angle $APM$ is right, then by the Pythagorean theorem $MT=\sqrt{P{T}^2+P{M}^2}.$

Combining $PT/MT=PM/MN$ (similar triangles with the common acute angle at $T$) gives

$MN=\frac{PM\cdot MT}{PT}=\frac{PM}{PT}\sqrt{P{T}^2+P{M}^2}.$

Why this matters

The subtangent and subnormal are the geometric outputs of the local jet $(A,B)$: they convert the algebraic ratio $-A/B$ into segments along the axis. Every classical conic theorem about tangents lands on a specific value of $-Bq/A$ — for instance, [[parabola|the parabola’s $PT=2x$]], or the constant subnormal $PN=c$ on the parabola (recovered here too). For curves where $(A,B)\ne (0,0)$, this method is complete; for points where both $A$ and $B$ vanish, see singular-points-by-jet.

Figures

Figures 55–60

Related pages

• chapter-13-on-the-dispositions-of-curves — chapter summary.
• tangent-by-translation — the underlying algorithm that supplies $A,B$.
• singular-points-by-jet — the case where $A=B=0$ and no subtangent is defined.
• parabola — §150’s $PT=2x$, recovered as Example I.
• ellipse — chapter 6’s tangent property, recovered as Example II.
• cubic-species-classification — the seventh-species cubic of Example III.