conic sections

Paths of planets, satellites and comets are beautiful curves called conic sections. They are obtained when a cone is intersected by a plane surface.

And also they have many other beautiful and interesting facts about them. For example they are such curves that distance from any point on the curve from a fixed point bears a constant ratio to distance from a fixed st line.

Different Sections Of A Double Cone

A cone cut by a plane a)Perpendicular to its axis of the cone gives a circle, if not through its vertex; through the vertex only a point is got.b) In an angle with the axis, gives an ellipse.c) Parallel to its slant height or generator gives a parabola.d) Parallel to its axis, gives two branches of a hyperbola.e) Through the axis, gives two straight lines.

A cone cut by a plane a)Perpendicular to its axis of the cone gives a circle, if not through its vertex; through the vertex only a point is got.b) In an angle with the axis, gives an ellipse.c) Parallel to its slant height or generator gives a parabola.d) Parallel to its axis, gives two branches of a hyperbola.e) Through the axis, gives two straight lines.

Fig 1 : Conic sections. a) plane section perpendicular to axis of cone – circle, if it is through the vertex , a point.

b) plane section with an angle with a plane perpendicular to axis of cone and not parallel to any of the generators – ellipse.

c) plane section parallel to a generator of the cone – parabola . d) plane section parallel to axis of cone not through it –

hyperbola,if through the axis of the cone – two straight lines.

Fig 1acircle

Fig1bellipse

Fig 1cparabola

Fig 1dhyperbola

ORBIT OF THE EARTH

We know the axis of earth always points towards the pole star. In fact that is the reason of changing seasons. If every point of the orbit is joined to the pole star, we get a cone ; not an odd shaped one, but a round smooth one called a right circular cone. Again the Sun is not placed at its center; but a t point called its focus, far away from the center.Planets and satellites move in elliptical orbitsComets who are seen periodically move around the Sun in vey long elliptical orbits.Some travelling comets at high speed are not captured by the Sun, they just bend around it once and pass away in parabolic orbits.Still some comets are in so high speed that they are deflected back from the Sun and turn away in hyperbolic orbits. This is some sort of collision.

We know the axis of earth always points towards the pole star. In fact that is the reason of changing seasons. If every point of the orbit is joined to the pole star, we get a cone ; not an odd shaped one, but a round smooth one called a right circular cone. Again the Sun is not placed at its center; but a t point called its focus, far away from the center.Planets and satellites move in elliptical orbitsComets who are seen periodically move around the Sun in vey long elliptical orbits.Some travelling comets at high speed are not captured by the Sun, they just bend around it once and pass away in parabolic orbits.Still some comets are in so high speed that they are deflected back from the Sun and turn away in hyperbolic orbits. This is some sort of collision.

The pole star

The Earth

The Sun

THE ‘ECCENTRICITY’ OF A CONIC SECTION Conic sections are curves such that ratio of distance from any point on it from a fixed point, called focus, bears a constant ratio ‘e’ or eccentricity to its distance from a fixed straight line, called directrix. The curve is a circle if e = 0, ellipse if e < 1, parabola if e = 1, hyperbola if e > 1, and two st lines if e = ∞.

The Circle and the ellipse are bound orbits. The moon’s orbit is almost circular orbit. Orbit of the earth is elliptical. Bodies with total energy negative are captured and move in bound orbits. This means potential energy of attraction exceeds kinetic energy which is always positive. Parabola is the orbit of a planet or comet critically captured; it takes a turn and escapes the other way, in ‘escape velocity’ corresponding to zero total energy of the escaping body. The body possesses total energy positive, i.e., it is in high velocity, outweighing attractive forces, it is deflected in hyperbolic orbit. This is some sort of collision, nay collision broader sense.

Equation of an ellipse

Let the eqn. of the circle be written in form ……………………………………..………….(a)

Transform the coordinates to x and y ,only by compressing Y coordinates by a factor b/a where b/a < 1;i.e.,x = X, and y = b .Y/a ; OR X = x and Y = a .y/b.………………………………….(b)The curve PMN is now changed to the curve QMN because of the transformation of coordinates to new coordinates x and y. A relationship between x and y shall be its equation. Since x and y are related to X and Y as in (b) and X and Y are related to each other as in (a), we can rewrite (a) as,

which is the only one relationship between x and y we sought for, and becomes the eqn. of the new curve ( called ellipse).

1b

y

b

x2

2

2

2

1b

y

a

x2

2

2

2

a

b

(x y)

O

XX’

Y

Y’

P

Q

RM N

2 2

2 21

x y

a a

2 2

2 21

X Y

a a

22

2 2

2 21

a yx ba a

2 2

2 21

x y

a b

A meaning Of eccentricity ‘e’

The circle has a centre and the ellipse has also a centre. But the centre of the ellipse is not that important as the centre of the circle. The ellipse has no such radial symmetry as in the case of circle. Instead, there are two other points of importance along the main axis or the major axis , one on each side of the center , each being called a focus. How far the focus is moved from the center so that the circle is flattened to become an ellipse is measured by its eccentricity. We explore below how the focus and eccentricity are related .In the eqn.of the ellipse , put b = fa or b/a = f, (evidently f < 1, as b < a). This reduces the eqn tox2f2 + y2 = a2f2…………….(a)

The left side looks like square of a distance (as we knew distance formula between the points (x, y) and (x1, y1), d =√{(x- x1) 2+ (y- y1) 2} and the form of the equation indicates us to make the expression a complete square so as to see what distance it might be. Put e2 = 1 – f2 (certainly we can, as f < 1) so that x2f2 becomes x2 (1 – e2 ), and a2f2 becomes a2 (1 – e2 ), Thus the eqn. becomes

x2 + a2e2 + y2 = a2 + e2x2

In further endeavour to complete squares, we add – 2aex to both sides and get (x – ae)2 + (y – 0)2 = e2(x – a/e)2………..(b)

Visualisation of e Spokes of a wheel being pulled by

two rods called directrices; the center splits into two points moved away from each other called foci.

The sum of focal distances from any point on the ellipse = 2a, the major axis.

Choose the origin midway between the two fixed points and choose the x-axis along the line joining the two fixed points. There is no loss of generality as we can shift the origin and the axes later as we please. Thus the coordinates of the two fixed points may be taken as (k, 0) and ( - k, 0) ; or as (ae, 0) and ( - ae, 0) if we denote k /a = e. Later on we would observe that this is really the eccentricity e. If (x, y) be the coordinates of the point , the sum of whose distances from (ae, 0) and ( - ae, 0) is 2a; we have, the steps are:

,

2 2 2 2( ) ( ) 2x ae y x ae y a

2 2 2 2 2 2 2 2 22 2 2 2 ( ) ( ) 4x a e y x ae y x ae y a

2

2 22 2 2 2 2 2 2 2 2 2 2 22 4x a e y a x a e y x a e

2 2 2 2 2 2 2a x e x a e y

2 2

2 2 21

(1 )

x y

a a e

2 2

2 21

x y

a b

Equation of any conic section in polar coordinates (origin at focus).

We would derive the equations of all conic sections in a uniform manner from the fact that the eccentricity e of the conic is the ratio of distance of any point on it from the focus to the length of perpendicular drawn from the point on the directrix. A chord through the focus parallel to the directrix is called latus rectum , LM in the figure and let its length be 2l. In the figure, let P be any point (x, y) or (r,) in polar coordinates,(OX being called the ‘initial line’ and , measured from this line anticlockwise being called ‘vectorial angle’ and r being called the ‘radius vector’).As such r/PD = e shall be our equation for all conic sections. Now, as L is a point on the conic, and LS = EF, and e =LF / EF = e. So that we have, EF = l/e. Draw ZZ’ a line through the vertex V of the conic, meeting PD at R.Now r/PD = e, or r = e.PD = e(PR + RD) = e(QF + FE)= e(r cos + l/e) = er cos + l Or, r = er cos + l , or, r - er cos = l, or, r(1 - e cos ) = l, or,

r = l/(1 - e cos ) …………………………..Eqn.(1)which is the equation of a parabola for e = 1, ellipse for e < 1, hyperbola for e > 1. The circle is a special case of ellipse where e = 0 and two straight lines ,a special case of hyperbola where e = .This would be evident by and by in later sections.If the curve is drawn to the left of the origin and the axes of coordinates x, y or r, are taken as usual , the eqn.(1) becomes

r = l/(1 + e cos ) ………………………….. Eqn.(1a)(observe that cos( - ) = - cos ),The curve is seen to be symmetrical about the x- axis as cos = cos ( - ), so that could be replaced by - .The semi-latus rectum l serves as a parameter for the conic section to describe the relationship between r and . The necessity of a parameter in addition to position of focus is actually to fix the position of the directrix as will be evident in the next paragraph. The directrix line could not have been fixed if only position of focus and eccentricity were given.

If the initial line XX’ is not the axis, and if the axis of the conic is inclined at an angle to it, then the geometry would be same if is replaced by (verify). So eqn.

becomes

Fig.12a : general conic section in polar coordinates

XX’

Y

Y’

L

M

F(0,0)

ZD

Z’D’

Q

R

Sr

V

P(x, y)

E

1 cos ......................................... (2)l

e Eqnr

Conic section in rectangular coordinates: (origin at directrix)

In rectangular coordinates, (see fig.12a ) , taking the directrix, DD’ as the y-axis, and taking the distance of the focus at a distance p from it, i.e., EF = p, and PD = EQ =x, we have,

PF2 /PD2 = e2 , PF2 = PD2e2 = e2x2, PF2 = FQ2 + PQ2 = (EQ – EF)2 + PQ2 =(x – p )2 + y2,Or, (x – p )2 + y2 = e2x2, p being called the focal parameter, being the distance from the focus to the directrix. Or, x2(1 – e2) + y2 – 2px +p2 = 0……………………Eqn.(3).

As FL = eSL = eEF, where FL = l and EF = p, we have l = pe………………………………………………. Eqn.(4)Semi-latus rectum = pe.


XX’

L

M

F(0,0)

ZD

Z’D’

Q

R

Sr

V

P(x, y)

E

Conic section in rectangular coordinates: (origin at focus)

If the origin is shifted to (p, 0), the old values of x should now be written X + p in terms of new coordinates and y = Y. So eqn. (2) becomes,(X + p)2(1 – e2) + Y2 – 2p(X + p) +p2 = 0or,X2(1 – e2) +2pX(1 – e2) + p2(1 – e2) + Y2 – 2pX - 2p2 +p2 = 0 or,,X2(1 – e2) + Y2 - 2e2pX – e2p2 = 0We can safely use x and y and forget about X and Y and write it asx2(1 – e2) + y2 - 2e2px – e2p2 =0Eqn(5)


XX’

L

M

F(0,0)

ZD

Z’D’

Q

R

Sr

V

P(x, y)

E

Conic section in rectangular coordinates: (origin at any point in general)

…………………………………………………… eqn(6) Is the equation of general conic section including circle, ellipse, parabola and hyperbola according as e = 0, e < 1, e =1, and e > 1. Focus is the point (h, k) and directrix is the st. line . The cases would be clear in the appropriate sections.Let the point V be the vertex at the origin (0, 0) and let the directrix DD’ be at a distance a , i.e., at the point G ( - a, 0). If the focus is at F, VF/VG = e, as VG = a, VF must be ae, so that F is at (ae, 0). If P(x, y) be any point on the conic, PF/PD = e, If PD cuts y-axis at R, or, PF/(PR +RD) = e, or, PF/(PR +VG) = e, or, PF/(VQ +VG) = e, or, r/(x + a) = e, From here we can go to r/(r cos + a ) = e, ………………………………eqn.(7)Or, ……………………………………………………………...eqn.(8)so , or …………………………………………………………………………..eqn(9)Or ………………………………………………………………eqn(10)

This ‘a’ must not be confused with semi major axis of the ellipse. Here it is the distance of the directrix from the vertex. Now l = LF = eSL = e(FV + VG) = e(ae +a);so …………………………..eqn(11)

22 2 2

2 2

( )( - ) ( )

Ax By Cx h y k e

A B

2 2 /(x + a) = ex y2 2 2 2=e (x + a)x y

2 2 2 2 2(1 ) 2 =0x e aex e a y

le

ae a

Derivation of Eqn of Ellipse in Cartesian Coordinates

The eqn. for general conic x2(1 – e2) + y2 – 2px +p2 = 0 becomes,x2(1 – e2) + y2 – 2 (a/e – ae )x +(a/e – ae)2 = 0 .

which is the equation of the ellipse. with origin at the directrix and focus at (a/e – ae,0).For, we already know that distance of the directrix of ellipse from its center is a/e , and distance of the focus from the center is ae. So p = a/e – ae.Only we have to shift the origin a/e to the right, i.e., replace x by x+ a/e So the eqn. taking origin at the centre of ellipse this eqn becomes,

(x+ a/e)2(1 – e2) + y2 – 2 (a/e – ae )( x+ a/e) +(a/e – ae)2 = 0 or, x2(1–e2)+y2 –2(a/e–ae)(x +a/e)+(a/e–ae)2+2x(a/e)(1– e2)+(a/e)2(1–e2) = 0 or, x2(1–e2)+y2 –2(a/e–ae)x–2(a/e–ae) a/e+(a/e–ae)2+

2x(a/e– ae)+ (a/e)2(1–e2) = 0 or, x2(1–e2)+y2 – 2(a/e)2(1 – e2) +(a/e–ae)2 +(a/e)2(1–e2) = 0or, x2(1–e2)+y2– (a/e)2(1 – e2) +(a/e–ae)2 = 0 , or, x2(1–e2)+y2– (a/e)2(1 – e2) +(a/e)2(1-e2)2 = 0or, x2(1–e2)+y2 - (a/e)2(1–e2) + (a/e)2(1-2e2 +e4) = 0or, x2(1–e2)+y2 - (a/e)2(1–e2 - 1+ 2e2 - e4) = 0 or, x2(1–e2)+y2 - (a/e)2e2(1 –e2) = 0 , or, x2(1–e2)+y2 – a2(1 –e2) = 0 or, , as before; or where as before.

2 2

2 2 21

1

x y

a a e

2 2 21b a e

Equation of parabola

This is an example of how we derive equations of conic section from focus directrix definition from the first principles.In the above figure and with the labels as before, A parabola is the curve such that any point on this curve, P(x, y) is at the same distance from the focus F as the distance from the directrix, PD. So PF = PD = QE.Next, if V(0, 0) is a point on the parabola, called its vertex, then , by above definition, VE = VF. Let us call VE = a, so that F is the point (a,0). Now, PF2 = r2 = FQ2 +QP2 = (x-a)2 +y2. Again PD2 = PR2 + RD2=(x+a) 2

Thus , or y2 = 4ax …………………………… ………..eqn(12)It the origin is transferred to the point ( - a , 0), or it is taken to the directrix, then x should be replaced by x – a , so that the equation to the parabola becomes,

y2 = 4a(x – a)…………………………. eqn(13a)It the origin is transferred to the point focus, ( a , 0), then x should be replaced by x + a , so that the equation to the parabola becomes,

y2 = 4a(x + a)…………………………. eqn(13b)


XX’

L

M

F

ZD

Z’D’

Q

R

Sr

V(0, 0)

P(x, y)

E

Equation of rectangular hyperbola

In the figure, APB be a curve where P(x, y) is any point on it such that OQ = x = RP and PQ = y = RO; and

xy = c2…………………..……………………..Eqn(14a)Such a curve is called a rectangular hyperbola characterized by the feature that the product of its distances from two fixed st.lines at right angles to each other , the coordinate axes in the present case is a positive constant say c2. The curve has two branches APB and A’P’B’ in the first and third quadrants If we move further to the right of any point P (x, y) the y- coordinate decreases and the x- coordinate increases in the inverse proportion so that xy remains constant. Down along the x- axis, the y- coordinate diminishes to 0 at large distance , so that the x- axis touches the curve at ∞. Similarly the curve touches the y – axis at ∞. A line touching a curve at ∞ is called an asymptote of the curve. Further, the branch A’P’B’ is reflection of the branch APB , not about any axis, but about the line y = - x shown as a dotted line. The fact may be verified by observing that the equation of this curve is unchanged by replacing x, y by – x and – y respectively. Further, it may be noted that the curve is symmetric about the line y = x , for interchanging x and y in the eqn makes no difference.

XO

Y

X’

Y’

P(x, y)

Q

R

A

B

A’

P’

B’

y= - x

R

R’

Fig.16, rectangular hyperbola

Equation of rectangular hyperbola

An analogy with the circle:In an ellipse, the sum of distances of any point on it is constant and equal to its major axis. Similarly, on this curve, the rectangular hyperbola, product of distances from two fixed st.lines is constant. ( its distance from y –axis being x and the distance from x – axis being y). The curve can be transformed by a change of axes , the new axes got from rotating the existing axes by 450 anticlockwise. The new set of axes, say X and Y form another rectangular Cartesian coordinate system about the same origin. If new coordinates of P be (X, Y) , we have, from principle of transformation of coordinates,x = X cos 450 – Y sin 450 and y = X sin 450 + Y cos450………….(b)Putting the values in (a),

(X cos 450 – Y sin 450)( X sin 450 + Y cos 450) = c2

or, (X /2 – Y /2)( X /2 + Y /2) = c2

or X2 – Y2 = 2c2 or, X2 – Y2 = a2

or, …………………..………………………….(14)

2 2

2 21

X Y

a a

XO

Y

X’

Y’

P(x, y)

Q

R

A

B

A’

P’

B’

y= - x

R

R’

Fig.16, rectangular hyperbola

Equation of hyperbola

Now the technique of compressing Y-coordinate may be applied. Let all Y- coordinates be compressed in the ratio ;so y’= Y,and x’ = X. Thus the eqn. reduces to

which may be re-written in usual form , for the only reason of sheer convenience as, ……………… Eqn.(15)We also can arrive at the ratio of focal distance to the directory distance of a point just in the manner we did for the ellipse. which can be regarded as the standard equation of a hyperbola. Just as the eqn. of ellipse is derived from the eqn. of a circle, the eqn. of hyperbola is derived from eqn. of a rectangular hyperbola in the same manner. Hence the rectangular hyperbola may be considered as a counterpart of a circle.

FF’

P(x,y)

D1

D1’

D2

D2’

O AA’

MM’

N’

N

L

L’

XX’

Y

Y’

Fig. 17 : Hyperbola

Q

P

Q

P’

Q’

b

a

2 2

2 21

x y

a b

2 2

2 2

' '1

x y

a b

b

a

TO SHOW THAT AN HYPERBOLA IS ACTUALLY A SECTION OF A CONE

Let a st.line VU revolve around a fixed st.line VG making a constant angle with it at V and generate a double cone as shown in the figure. Let a plane parallel to VG and perpendicular to the plane of VUW cut the double cone in a curve in two branches LMN and L’M’N’ , M and M’ being two points on the cone. Let VC = b , be perpendicular to MM’. Set up a rectangular Cartesian coordinate system at C , midpoint of M’M and MM’ being the x-axis and YY’ being y-axis in the intersecting plane. Take any point P on the curve of intersection and draw a perpendicular PR onto MM’ and extend it until it meets the curve at N. Coordinates of the point P are x = CR and y = PR. Take a plane containing PR and perpendicular to the plane VST . this plane intersects the cone in a circle PST having its centre at O and radius OP = OS = OT = r.

C(0, 0) L

M L

M’ L

b L

S L

T L

P

O L

R L

N L

r L

G L

U L

W L

y

x L

T’ L

S’ L

L’ L N’

L

x L

b L

L L

Y L

Y’

X’

X L

V L

900

a L

a L

Fig. 18 : A hyperbola is really a section of a cone

TO SHOW THAT AN HYPERBOLA IS ACTUALLY A SECTION OF A CONE

This can be proved from congruence of the two triangles VSO and VOT, having the side VO common, and two equal angles SVO = = OVT and having a right angle each. OVCR can be proved to be a rectangle as three of its angles are right angles, so that OR = VC = b. Now PR is in the intersecting plane and OR is in the plane of the circle PST and the two planes are perpendicular to each other. So OR PR and it follows that OP2 = OR2 + PR2.Or, r2 = b2 + y2…………………….(a)

C(0, 0) L

M L

M’ L

b L

S L

T L

P

O L

R L

N L

r L

G L

U L

W L

y

x L

T’ L

S’ L

L’ L N’

L

x L

b L

L L

Y L

Y’

X’

X L

V L

900

a L

a L

Fig. 18 : A hyperbola is really a section of a cone

Continued from the previous slide

As OP = r, OR = b and PR = y..The relationship between x and y shall be the eqn. to the curve of intersection we require, which can be obtained from this eqn.(a) r in terms of x. We immediately observe that r = OS = OV tan and OV = CR = x. Hence we get, x2tan2 = b2 + y2………………………………...(b)

is the required equation to the curve of intersection. If we denote the length CM = a, observe that CMV = MVO = , so that tan = b/a. Now eqn. (b) becomes, x2b2 = a 2b2 + a2y2 or,

…………….( c)which is standard equation to a hyperbola . Note that ‘a’ is actually seen to be its semi-major axis and ‘b’ is equal to its semi-minor axis, though it is not in the plane of the curve. The value of e may be obtained from ,

…………………………………(d)

which is always greater than 1 for any given acute angle . For a different value of e we need a different value of , or in other words, need a different cone altogether. From one cone, we get hyperbolas all of same e value i.e., sec2. This is because either we have to choose a different cone or different values of a and b; to get different hyperbolas from the same cone, i.e., the plane of intersection must be different.

2 2

2 21

x y

a b

22 2 2

21 1 tan secb

ea

Continued from the previous slide

This is a meaning of e for the hyperbola.It may be further noted that in the eqn. of an ellipse ,or , where , e is supposed to be less than 1, i.e., b < a. If we make e > 1 , e becomes imaginary and we do not get a hyperbola in place of an ellipse, as b becomes imaginary; not evenif we rewrite the same equation as . On the other hand, if we make b > a in , we get only another ellipse with its major and minor axes interchanged. The hyperbola shown in the figure of this article is represented by ; where b is the length of the perpendicular from vertex of the cone onto the plane of intersection , i.e., onto the plane of the hyperbola. But as it is the standard eqn. of hyperbola, b is supposed to be less than a , and it is very much the semi-minor axis, which is supposed to be in the plane of the hyperbola. Thus the length of the perpendicular from vertex of the cone happens to be equal to the semi-minor axis, the two being perpendicular to each other. This is as if the minor axis has been rotated through /2 or having been multiplied by i = (-1) , perhaps because b2 in the equation of the ellipse is replaced by - b2 in contrast to the case of ellipse. Then which is always greater than 1.

2 2

2 21

x y

a b

22 2 2

21 1 tan secb

ea

2 2

2 2 21

(1 )

x y

a a e

2 2 2(1 )b a e

2 2

2 2 21

( 1)

x y

a a e

2 2

2 21

x y

a b

The general eqn. of 2nd degree in two variables represents a conic.

The General eqn. of 2nd degree ax2 +2hxy + by2 + 2gx +2fy +c = 0……….…eqn(16)By a suitable transformation of co-ordinates , the xy-term may be made to vanish. Suppose the axes are turned through an angle so that, x is replaced by x cos -y sin and y is replaced by x sin + y cos ,so that (16) becomes a(x cos - y sin )2 + 2h(x cos - y sin )( x sin + y cos )+ b(x sin

+ y cos )2+2g(x cos - y sin ) + 2f(x sin + y cos ) + c = 0.The coefficient of 2xy term is h(cos2 -sin2 ) – (a – b) cos sin ,which is 0 if, 2h cos 2 = (a – b) sin 2 , or tan 2 = 2h/(a – b).For any real value of h, a and b, can always be found, so that the eqn. can always be got rid of the xy-term (second term).Even for a = b, cos 2 = 0 and = /4 if we put a = b from the beginning. And thus can be written in the formAx2 +By2 + 2Gx +2Fy +C = 0…………………………………………………………….(17)Of course c = C as the term does not involve the variables only whom we have modified. Thus we can take (17) as general equation of second degree in two variables without any loss of generality.This equation can be written as follows by completing squares ……………………………………...(18)

if A 0 and B 0.

2 2 2 2

2 2( )

G F G FA x B y C K say

A B A B


If the origin is shifted to the point , eqn.(17) can be written as , ..................................... (18)

(to keep the number of symbols minimum, we write x and y instead of new symbols, though x and y in (18) are different from x and y in (17)or simply transfer the origin to )The eqn. represents an ellipse , real if K/A and K/B both are positive, or imaginary one , if K/A and K/B both are negative, their roots are imaginary. This shows that a second degree eqn in two variables Ax2 +By2 + 2Gx +2Fy +C = 0 in which there is no xy-term shall be equation of an ellipse if A and B are of same sign as that of . Eqn.(18) represents a hyperbola if one of K/A and K/B is negative. This shows that a second degree eqn in two variables Ax2 +By2 + 2Gx +2Fy +C = 0 in which there is no xy-term shall be equation of an hyperbola if A and B are of opposite signs.

2 2

1x yK KA B

,G F

A B

,G F

A B

2 2

2 2

G FC

A B


If A = B, this represents a circle , origin at and radius ………………...(19)

and only a point , trivial case of circle if K = 0. If one of A, B, is 0 , say A = 0 and B 0, from eqn.(18) we get,

By2 + 2Gx +2Fy +C = 0……………………………………………..…..(20)By completing the square,

Or, By2 +2Gx = 0 or, ……………………………………..(21)

which represents a parabola if the origin is shifted to its branches towards negative side of the x-axis if G/A is positive and vice versa.If G = 0 along with A = 0, from eqn.(17) we get,In other words, a second degree eqn in two variablesAx2 +By2 + 2Gx +2Fy +C = 0 in which there is no xy-term shall be equation of a parabola, if A = 0 , G 0.

By2 + 2Fy + C = 0 or ………………………………………….(22)which represents two straight lines parallel to each other and parallel to x-axis. The straight lines are real if F2 – BC is positive, or else, imaginary.

K Kor

A B

2 2

2 02 2

F C FB y G x

B G BG

2 2Gy x

B

2

,2 2

C F F

G BG B

2F F BCy

B

Summary

The general conic represents the following curves under respective conditions.

What curve Under what condition

Ellipse

Parabola

Hyperbola

Circle a = b and h =0

Rectangular hyperbola a + b =0

Two st lines, real or imaginary Δ =0, where Δ =abc +2fgh – af2 – bg2 – ch2

Two parallel st lines Δ =0 and

2h ab

2h ab

2h ab

2h ab

Education

conic sections