tangent-properties-conic

Tangent Properties of a Conic (Newton’s Lemmas)

Summary: Take any diameter $AB$ of a conic with conjugate $ECF$, and draw tangents $AK,BL$ at the diameter’s two endpoints (both parallel to $ECF$, by conjugate-diameters §111). For any third tangent $MT$ at any point $M$ of the curve, intersecting the two given tangents at $K$ and $L$, the rectangle property reads $AK\cdot BL=C{E}^2$ (§122). Adding a fourth tangent $\lambda \mu \omega$ further intersecting at $\lambda ,\omega$ on $IL,KL$, one obtains $\frac{I\ell }{I\lambda }=\frac{Ko}{K\omega }$ (§123). And any straight line cutting two of the tangents in a fixed ratio cuts the third in the same ratio (§124). Euler concludes: “These are the main properties of conic sections from which NEWTON found the solution to many important problems in his Principia” (source: chapter5, §122).

Sources: chapter5 §§121–125, figures26-29 (figures 28, 29), figures30-32 (figure 30)

Last updated: 2026-04-25

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Setup (§121, figure 28)

Take any diameter $AB$ of the conic, with center $C$ at its midpoint and conjugate diameter $ECF$. The tangents $AK,BL$ at the endpoints $A,B$ are both parallel to $ECF$ (a consequence of the conjugate-pair structure — conjugate-diameters §111).

Pick any point $M$ on the curve and draw the tangent $MT$ at $M$, extending it until it meets $AK$ at $K$ and $BL$ at $L$. Drop the ordinate $MP$ from $M$ to the diameter $AB$, with $MP$ parallel to $ECF$ (so $MP$ is parallel to both $AK$ and $BL$).

The first ratio: $CP\cdot CT=C{A}^2$ (from §118)

From the tangent-construction theorem of conjugate-diameters §118, one has $\frac{CP}{CA}=\frac{CA}{CT},\text{i.e.,}C{A}^2=CP\cdot CT.$ Equivalently $CP\cdot CT=C{B}^2$ since $CA=CB$. By componendo / dividendo on the proportion $CP/CA=CA/CT$:

• $\frac{CP}{AP}=\frac{CA}{AT}⟹AT=\frac{CA\cdot AP}{CP}$.
• $\frac{CP}{BP}=\frac{CA}{BT}⟹BT=\frac{CA\cdot BP}{CP}$.

So $PT=AT-AP=\frac{CA\cdot AP}{CP}-AP\cdot \frac{CP}{CP}=\frac{AP(CA-CP)}{CP}\cdot (\text{etc.})⟹PT=\frac{AP\cdot BP}{CP}.$

(Algebra works out because $CA-CP=AP$ on one side and $CA+CP=$ etc. on the other; sign conventions follow Euler.)

Tangent intercept ratios (§121)

By similar triangles $△ATK\sim △BTL$ (with bases on the same diameter and parallel sides $AK,BL$): $\frac{AT}{BT}=\frac{AK}{BL}.$ But also $\frac{AT}{BT}=\frac{AP}{BP}$ from the formulas above (dividing them). Hence: $\frac{AK}{BL}=\frac{AP}{BP}.$

Furthermore, by similar triangles: $\frac{AT}{PT}=\frac{AK}{PM},\frac{BT}{PT}=\frac{BL}{PM}.$ Combining: $AK=\frac{CA\cdot PM}{BP},BL=\frac{CA\cdot PM}{AP}.$

$AK\cdot BL=C{E}^2$ (§122)

Multiply: $AK\cdot BL=\frac{C{A}^2\cdot P{M}^2}{AP\cdot BP}.$

But by the chord-rectangle-property (specialized: at the diameter $AB$, the chord at abscissa $CP$ from the center has half-length $PM$, and the “axis-chord” is $AB$ with $PE\cdot PF\to AP\cdot BP$), $\frac{P{M}^2}{AP\cdot BP}=\frac{C{E}^2}{C{A}^2}.$ (This is the conjugate-diameter form of the rectangle ratio: $\delta /\zeta$ has become $C{E}^2/C{A}^2$ after referring everything to the principal-pair-like center frame.)

Therefore: $AK\cdot BL=C{A}^2\cdot \frac{C{E}^2}{C{A}^2}=C{E}^2.$

From this there follows the outstanding property $AK\cdot BL=C{E}^2$. (source: chapter5, §122)

The product of the lengths the third tangent cuts off on the two end-tangents is the square of the conjugate semidiameter — and is therefore the same for every choice of $M$.

Equivalent forms (§122)

Several immediate corollaries: $AK=CE\sqrt{\frac{AP}{BP}},BL=CE\sqrt{\frac{BP}{AP}},\frac{AP}{BP}=\frac{A{K}^2}{C{E}^2}=\frac{C{E}^2}{B{L}^2}=\frac{KM}{ML}.$

The last identity, $\frac{AK}{BL}=\frac{KM}{LM}$, says that the point $M$ on the tangent line $KL$ divides $KL$ in the same ratio as $\frac{AK}{BL}$ — i.e., the contact point divides the third tangent in the same ratio that the third tangent divides the two end-tangents (in opposite senses). This is one of the key projective lemmas Newton uses.

Two further tangents and four-tangent ratios (§123)

Take a fourth point $m$ on the curve with tangent $kml$ meeting $AK$ at $k$ and $BL$ at $\ell$. The same identity gives $Ak\cdot B\ell =C{E}^2$. Hence: $\frac{AK}{Ak}=\frac{B\ell }{BL}.$

Now if the two tangents $KL,k\ell$ meet at $o$, then by similar triangles $\frac{AK}{B\ell }=\frac{Ak}{BL}=\frac{Kk}{L\ell }=\frac{ko}{\ell o}=\frac{Ko}{Lo}.$

These are the main properties of conic sections from which NEWTON found the solution to many important problems in his Principia. (source: chapter5, §122–123)

The four-tangent harmonic identity (§§123–124, figure 30)

If the tangent $LB$ is extended to a point $I$ such that $BI=AK$, then $I$ is the point where the tangent on the other side, parallel to $KL$, intersects $LB$ — and likewise $K$ on the other side. So tangents $BL,ML$ extended to $I,K$ create a configuration where, for any third tangent $\ell mo$ at points $\ell ,o$: $\frac{IB}{I\ell }=\frac{Ko}{KL}.$

Hence $IB\cdot KL=I\ell \cdot Ko$. Adding a fourth tangent $\lambda \mu \omega$ intersecting $IL,KL$ at $\lambda ,\omega$: $IB\cdot KL=I\lambda \cdot K\omega ⟹\frac{I\ell }{I\lambda }=\frac{Ko}{K\omega }.$

If $\ell ,\omega$ are bisecting points (each cutting their respective tangent in the same fixed ratio $m:n$), then the line through them cuts $IK$ in the same ratio. In particular: the line through the midpoints of $\ell \omega$ and $\lambda o$ bisects $IK$ — and therefore passes through the center $C$ of the conic (because $C$ is the midpoint of $IK$ when $IB=AK$ is set to make the configuration symmetric about $C$).

If the lines $\ell \omega$ and $\lambda o$ are bisected, the straight line through the two bisecting points will also bisect the straight line $IK$. For this reason it will also pass through the center $C$ of the conic section. (source: chapter5, §123)

Geometric proof of the ratio invariance (§124, figure 30)

Euler’s geometric proof reduces $\frac{Q\ell }{R\omega }$ to the ratio $\frac{m}{n}$ (the prescribed division ratio of $\ell \omega$) by the law of sines applied twice in triangles $Q\ell \lambda$ and $Ro\omega$: $\frac{\sin Q}{\sin R}=\frac{m/(\ell \lambda )}{n/(o\omega )}.$

Cross-multiplying and using $\frac{\ell \lambda }{o\omega }=\frac{I\lambda }{Ko}$, $\frac{Q\ell }{R\omega }=\frac{Q\lambda }{Ro},\frac{HI}{HK}=\frac{m}{n}=\frac{\lambda m}{mo}=\frac{\ell n}{n\omega }.$

The conclusion (§124):

If the straight line $\ell \omega$ divides both $\ell \omega$ and $\lambda o$ in the ratio $m:n$, the straight line through the two division points divides the straight line $IK$ in the same ratio. (paraphrased from chapter5, §124)

Why these are “Newton’s lemmas”

In the Principia, Newton derives Kepler’s equal-area law and the inverse-square force from the conic’s tangent structure. The pivotal property he uses repeatedly is precisely §122’s $AK\cdot BL=C{E}^2$, applied to areas of inscribed triangles and tangent-quadrilaterals. The §123–124 ratio invariances are how Newton transports the area-ratio argument across different chords of the orbit.

Euler’s reference to Newton in §122 is no afterthought — the chapter is in part a clean algebraic derivation of the synthetic conic geometry Newton used. By the end of §124 Euler has, in fewer than a dozen pages, produced the full toolkit Newton spent the Principia’s opening sections developing.

Connection to the chord-rectangle property

The tangent property $AK\cdot BL=C{E}^2$ is the limiting case of the chord-rectangle-property’s tangent-rectangle form $\frac{P{C}^2}{PM\cdot PN}$ being constant (§94), specialized to the configuration where $A,B$ are the endpoints of a diameter and the tangents at $A,B$ meet the third tangent at $K,L$. Both rest on the algebraic fact that the product of roots of the conic’s quadratic-in-$y$ equation factors cleanly (§92).

Figures

Figures 26–29

Figures 30–32

Related pages

• chapter-5-on-second-order-lines
• chord-rectangle-property — the tangent-rectangle theorem $P{C}^2\propto PM\cdot PN$ that specializes to this property
• conjugate-diameters — tangents at the endpoints of a diameter are parallel to the conjugate diameter
• principal-axes-and-foci — applying these tangent ratios in the orthogonal-pair frame
• diameter-of-conic