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The Squares of the Ordinates of the Axis are to each other

as the Rectangles of their Abscisses. LET AVB be a plane passing through the axis of the cone; AGIH another section of the cone perpendicular to the plane of the former; AB the axis of this elliptic section; and PG, HI,

L dinates perpendicular to it. Then it

M/H. will be, as FG2: HI::AF. FB: AH . HB.

For, through the ordinates FG, HI, draw the circular sections KGL, MIN, parallel to the base of the cone, having KL, MN, for their diameters, to which FG, HI, are ordinates, as well as to the axis of the ellipse.


Now, by the similar triangles AFL, AHN, and BFK, BHM,


it is AF : AH :: FL : HN,
and FB : HB :: KF: MH;

hence, taking the rectangles of the corresponding terms, it is, the rect. AF . FB : AH. HB :: KF FL : MHHN.

But, by the circle, KF. FL = FG2, and MH . HN = 1; Therefore the rect. AF. FB : AH . HB :: FG? : HI?. Q. E. D.

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For, by theor. 1, AC. CB : AD . DB:: ca? : DE2;
But, if c be the centre, then AC CB = AC’, and ca is the

semi-conjugate. Therefore

AC : AD. DB : : acé : DER; or, by permutation, ac : ac? :: AD DB : DE; or, by doubling, AB? : ab? :: AD. DB : DE?.

R. E, D, ab? Coral. Or, by div. AB: :: AD. DB or CA? CD : DE,

AB that is, AB :p :: AD


CD : De°;

ab? where p is the parameter

by the definition of it.

That is, As the transverse,
Is to its

So is the rectangle of the abscisses,
To the square of their ordinate.

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As the Square of the Conjugate Axis :
Is to the Square of the Transverse Axis ::
So is the Rectangle of the Abscisses of the Conjugate, or

the Difference of the Squares of the Semi-conjugate and

Distance of the Centre from any Ordinate of that Axis : To the Square of their Ordinate.

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For, draw the ordinate ED to the transverse AB. Then, by theor. 1, ca’: CA’ :: DE’: AD. DB or CA’ – CD',

ca? : CA? :: cd? : CA de?. But

- cal : CA :: ca?: CA, theref. by suber. ca? : CA? : : ca? cd? or ad . db : de?.

Q. E, D.


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DB =

Corol. 1. If two circles be described on the two axes as diameters, the one inscribed within the ellipse, and the other circumscribed about it, then an ordinate in the circle will be to the corresponding ordinate in the ellipse, as the axis of this ordinate, is to the other axis.

That is, ca : ca :: DG : DE,

and ca : CA :: dg : de. For, by the nature of the circle, AD

DG?; theref. by the nature of the ellipse, CA’ : ca’ :: AD. DB or DG? : DE,

or ÇA: ca :: DG : DE. In like manner

ca : CA :: dg : de. Also, by equality, DG : DE or cd :: de or dc : dg.

Therefore cgG is a continued straight line. Corol: 2. Hence also, as the ellipse and circle are made up of the same number of corresponding ordinates, which are all in the same proportion of the two axes, it follows that the areas of the whole circle and ellipse, as also of any

like parts of them, are in the same proportion of the two axes, or as the


of the diameter to the rectangle of the two axes; that is, the areas of the two circles, and of the ellipse, are as the square of each axis and the rectangle of the two; and therefore the ellipse is a mean proportional between the two circles.


The Square of the Distance of the Focus from the Centre,

is equal to the Difference of the Squares of the Semi

axes ; Or, the Square of the Distance between the Foci, is equal to the Difference of the Squares of the two Axes.


That is, CF= CA? - ca,

or Ff2 = AB? ab2.

For, to the focus F draw the ordinate FE; which, by the definition, will be the semi-parameter. Then, by the nature of the curve

CA? : ca' :: CAP CF2 : FE”; and by the def. of the para. CA? ; ca ::

: FE?; therefore

ca? -- CAP and by addit, and subtr.

CA or, by doubling,


CF? ; ca> ;


ff? ==

AB? · ab?.
H 2


Q: E. D.

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Corol. 1. The two semi-axes, and the focal distance from the centre, are the sides of a right-angled triangle cha; and the distance fa from the focus to the extremity of the conjugate axis, is = Ac the semi-transverse.

Corol. 2. The conjugate semi-axis ca is a mean proportional between AF, FB, or between af, fb, the distances of either focus from the two vertices.

For ca? = CA? CE? = CA + CF. CA CF = AF.FB.


The Sum of two Lines drawn from the two Foci to meet

at any Point in the Curve, is equal to the Transverse Axis.

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For, draw AG parallel and equal to ca the semi-conjugate; and join cg meeting the ordinate de in h; also take ci 2 4th proportional to CA, CF, CD. Then, by theor. 2, CA2:AG:: CA: - CD2 : DE; and, by sim. tri. CA?: AG? :: CA? CD2: AGP DH”; consequently DE? = AGP DH? = ca? - DH?. Also FD = CF 02. CD, and Fd’ = CFP 2CF. CD + cd? ; And, by right-angled triangles, FE’ = FD? + DE”; therefore we? = CF? + ca’ – 2CF . CD + CD? But by theor. 4, CF + ca’ = ca', and by supposition, 2CF. CD = 2CA . CI; theref. FE? = CA? 2CA . CI + CD? Again, by supp. CA? : CD’:: CF? or ca’ – AG? : ci”; and, by sim. tri. CA?: CD?:: CA

AGP : CD? – Dho; therefore CI? CD consequently FE= CA? 2CA . CI + ci?. And the root or side of this square is Fe = CA In the same manner it is found that fe = c + c = BI. Conseq. by addit. Fe + fe = a1 + BI = AB. Q. E. D.






Corol. 1. Hence cı or CA - FE is a 4th proportional to CA, CF, CD.

Corol. 2. And fe FE = 2c1; that is, the difference between two lines drawn from the foci, to any point in the curve, is double the 4th proportional to CA, CF, CD.

Corol. 3. Hence is derived the common method of describing this curve mechanically by points, or with a thread, thus :

In the transverse take the foci F,f, and any point 1. Then with the radii AI, BJ, and centres F, f, describe arcs intersecting in E, which will be a



FI point in the curve. In like manner, assuming other points 1, as many other points will be found in the curve. Then with a steady hand, the curve line may

be drawn through all the points of intersection E.

Or, take a thread of the length AB of the transverse axis, and fix its two ends in the foci F, f, by two pins. Then carry a pen or pencil round by the thread, keeping it always stretched, and its point will trace out the curve line.


THEOREM VI, If from any Point i in the Axis produced, a Line il be

drawn touching the Curve in one Point L; and the Ordinate im be drawn; and if c be the Centre or Middle of AB: Then shall cm be to ci as the Square of AM to the Square of ai.


That is,

CM :CI :: AM? : AI?.


For, from the point i draw any other line ish to cut the curve in two points E and h; from which let fall the perpendiculars ED and HG; and bisect DG in K.

Then, by theo. 1, AD. DB : AG · GB: D E : G H ́, and by sim. triangles,


IG2 : : DE: GH? ; theref. by equality, AD , DB : AG . GB :: ID: 1G2. But DB = CB +CD=AC + D = AG +'DC CG= 2CK + AG, and GB = CB - CG = AC CG I AD + DC – CG = 20K + AD; theref. AD. 2cK + AD. AG:AG . 2CK + AD. AG :: ID2:16?, and, by div. DG . 20K : 162 ID? or DG . 2IK :: AD , 2cK +

AD. AG : ID,


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