Ellipse-1 Basics

LOCUSLOCUSLOCUSLOCUSLOCUS 1

Mathematics / Ellipse

01. Introduction

02. Tangents

03. More on Tangents and Chords

04. Normals

CONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOTESTESTESTESTES



Parabolas, which we studied in the last chapter, are conic sections with eccentricity e = 1. In this section, we studyellipses, which have e < 1. This means, by definition, that an ellipse is the locus of a moving point such that the ratioof its distance from a fixed point to its distance from a fixed line is a constant less than unity. In fact we will soondiscover that an ellipse has two foci and two directrices.

We now start by first writing down the basic equations representing an ellipse.

Let F represent the focus of an ellipse and L be its directrix. Let Q be any point on the ellipse.

Q

F

N

L

If the eccentricity of this ellipse is given to be e, we must have

Distance of from Distance of from the line

Q F QF eQ L QN

Draw a line through F perpendicular to the directrix L, which meets the directrix at M. On this line, we can alwaysfind two points, say P and P', which divide the line segment MF internally and externally respectively in the ratio1 : e.

y

xM

Directrix L

P F P'

We can always find two pointsand ' such thatP P

PFPM = P F'

P M' = e

Thus, by definition, P and P ' will both lie on the ellipse.

Section - 1 INTRODUCTION



We will now introduce co-ordinates into this geometry. We assume the x-axis to coincide with the horizontal lineand the origin to be at the mid-point of PP'. The co-ordinates of P and P' can be assumed to be (–a, 0) and(a, 0):

y

xP -a, ( 0)

FP -a, '( 0)M

L

SincePF ePM and ,P F e

P M we have

( )PF P F e PM P M

( )PP e PM P M

2 (2 )a e PM PP

2 2ePM ae

(1 )aPM ee

This gives (1 )PF ePM a e and .aOM OP PMe

Thus the co-ordinates of F are (–ae, 0) and that of

M are , 0 .ae

Till now we have assumed F to be the focus of the ellipse and L to be the directrix. However, by the symmetryinherent in Fig -3, you might be able to infer that there should exist another focus F ' and directrix L ', to the rightof O as shown below:

y

xM OP F

(– ,0)a (– ,0)aeF '

( ,0)aeP '

( ,0)aM 'ae ,0)(-a

e ,0)(

L L'

That there exist another focus and directrix will become evident once we obtain the equation of the ellipse.



To find the equation of the ellipse we assume the co-ordinates of any point Q lying on the ellipse as (x, y):

y

xF(– ,0)ae F ' ( ,0)ae-a

e ,0)(

L L'

NQ x, y( )

N '

ae ,0)(

We thus have

QF eQN

2 22

2

( )x ae y eaxe

2 2 2 2 2 2 22 2x y a e aex e x a aex

2 2 2 2 2(1 ) (1 )x e y a e

2 2

2 2 2 1(1 )

x ya a e

2 2

2 2 1x ya b where 2 2 2(1 )b a e

This is the equation of the ellipse; any point (x, y) on the ellipse must satisfy this equation.

The symmetric nature of this equation should convince you that there does indeed exist another focus F ' andanother directrix L '. In fact, you are now urged to verify the truth of this explicitly by proving that for any pointQ(x, y) lying on the ellipse as in Fig - 5, we always have

QF QF eQN QN



We finally plot the ellipse taking help of the equation obtained.

y

xP '

L L'

NQ

N '

M 'F 'F OP

R

M

R'

The co-ordinates of R and R' can be obtain by substituting x = 0 in the equation of the ellipse obtained :

y b

Thus, (0, )R b and (0, )R b .

PP', whose length equals 2a, is termed the of the ellipse, while RR ', whose length equals 2b, is termedthe of the ellipse. The names are self-explanatory. The coordinates of F and F ' have already been

mentioned : (–ae, 0) and (ae, 0) so that FF ' = 2ae. Also, 2 .aMMe

We now note a very important fact about any ellipse from Fig - 6. Since Q lies on the ellipse, we have by definition,

QF QF eQN QN

( )QF QF e QN QN

( )e NN

( )e MM

2aThis means that for any point on the ellipse, the sum of its distances from the two foci is a constant, and equal to themajor axis of the ellipse. In fact, this property itself is sometimes used to define an ellipse:

An ellipse is a locus of a moving point such that the sum of its distances from two fixed points and , i.e., + , is a constant. and are termed the foci of the ellipse.

QF F QF QF F F

The two definitions of the ellipse, one through ‘eccentricity’ and one through the ‘constant sum of distances’, areequivalent.



There is a very nice way to plot an ellipse very precisely. Fix two pegs on a board at a distance l, and tie a stringof length L > l between these two pegs. Using your pen, stretch this string away from the pegs, so that it becomestaut :

l l

Now always keeping the string taut, mark out a complete revolution on the paper with the pen. You’ve just madean ellipse ! This is because keeping the string taut ensures that the sum of the distances of the tip of your pen fromthe two foci is always constant, equal to L.

l

Move the pen while keeping the string taut

The pen traces out an 'ellipse'

Can you calculate the eccentricity of this ellipse we just made? If we compare this ellipse with the standard formobtained earlier, we have

2ae l

2a L

leL

We now discuss some terminology related to an ellipse which we’ve already seen for the case of parabolas.



Let 2 2

2 2 1x ya b

be the ellipse and F, F ' be the two foci (refer to fig - 6). The focal

distances of Q are simply the two distances QF and QF .

aQF eQN e x a exe

aQF eQN e x a exe

Thus, the two focal distances of any point (x, y) are (a + ex) and (a – ex). The sum is2a, as expected.

: This is the chord passing through any of the two foci and perpendicular to the majoraxis. To evaluate the length of the latus-rectum, we must evaluate the y-coordinates ofthe two extremities of the latus rectum. In the equation of the ellipse

2 2

2 2 1x ya b

we substitute x ae ( the x-coordinates of the two foci). Thus

22 2

21 xy ba

2 2 2(1 )a e

2(1 )y a e

2b

a

Thus, the length l of the latus-rectum is

22bla

Find the lengths of the major and minor axes and the foci for these ellipses:

2 216 25 400x y

2 24 2 0x y x

2 225 16 400x y

Example – 1



The given equation can be written in the form of the standard equation of an ellipse :2 216 25 1

400 400x y

2 2

125 16x y

2 2

2 2 15 4x y

Thus, comparing with the standard form, we have

a = 5, b = 4The major axis is of length 2a = 10 and the minor axis is of length 2b = 8.The eccentricity of the ellipse is

2

2

315

bea

Thus, the two foci are at ( , 0),ae i.e., at ( 3, 0).

We first rearrange the equation of the ellipse to standard form :2 24 2 1x y x

2 2( 1) 4 1x y2 2

2 2

( 1) 111 ( )2

x y

Instead, of x, we have x –1. This means that the center of the ellipse is at (1, 0) instead of (0, 0).As in the unit on parabola, we can use a translation of the axes (refer to Parabolas) :

1,X x .Y y

so that the equation of the ellipse in the X-Y system is2 2

22 11 1

2

X Y

Since a = 1 and 1 ,2

b the major and minor axes are 2 and 1 respectively, while the eccentricity

is 1 31 .4 2

e Thus, the two foci are at (in the X-Y system):

3 ,02

In the original x-y system, we use the reverse transformation 1, ,x X y Y so that the fociin the original system are at

31 ,02



2 225 16 400x y

2 2

116 25x y

2 2

2 2 1(4) (5)x y

Here, we see that b = 5 is greater than a = 4, which means that the major axis will lie not along thex-axis but along the y-axis, and obviously, since the foci lie on the major axis, the foci will also lieon the y-axis. The major and minor axes are of lengths 2b = 10 and 2a = 8 respectively. The

eccentricity is 2

21 aeb

(a and b get interchanged ) 3 .5

Thus, the two foci lie at (0, ),be

i.e. at (0, 3).

y(0 5),

F1

F2

(4 0),(-4 0),

(0 5),-

x

Since the major-axis is along the -axis, the twofoci also lie along the -axis. Their coordinates are

yy

F = 1 (0, 3)F = 2 (0, –3)

and

The length of the latus rectum in this case will be given by 22a

b and will be equal to

325



We pause briefly and summarize whatever we’ve covered uptill now.

SUMMARY

BASICS ON ELLIPSE

Definition – 1

An ellipse is the locus of a moving point such that the ratio of its distance from a fixed point to its distance from a fixed line is a constant less than unity. This constant is termed the eccentricity of the ellipse. The fixed point is the focus while the fixed line is the directrix. The symmetrical nature of the ellipse ensures that there will be two foci and two directrices.

Definition – 2

An ellipse is the locus of a moving point such that the sum of its distances from two fixed points is constant. The two fixed points are the two foci of the ellipse. To plot the ellipse, we can use the peg-and-thread method described earlier.

STANDARD EQUATION

2 2

2 2 1x ya b

Vertices Foci Major axis Minor axis Directrices Eccentricity e Latus-recutm Focal distances of (x, y)

(a, 0) and (-a, 0) (ae, 0) and (-ae, 0) 2a (along x-axis) 2b (along y-axis)

anda ax xe e

2

21 ba

22b

a

a ex

(0, b) and (0, -b) (0, be) and (0, -be) 2b (along y-axis) 2a (along x-axis)

andb by ye e

2

21 ab

22a

b

b ey



And lastly, if the equation of the ellipse is

2 2

2 2

( ) ( ) 1x ya b

instead of the usual standard form, we can use the transformation X x and Y y (basicallya translation of the axes so that the origin of the new system coincides with ( , ). The equation thenbecomes

2 2

2 2 1X Ya b

We can now work on this form, use all the standard formulae that we’d like to and obtain whatever itis that we wish to obtain. The final result (in the x-y system) is obtained using the reverse transformationx X and .y Y

Find the equation of the ellipse with centre at the origin, foci at ( 1, 0) and 1 .2

e

Let the major axis be of length 2a. The distance between the two foci, 2ae, is equal to 2 for thisexample.

2 2ae

1 2ae

Thus,

2(1 ) 3b a e

The equation of the ellipse is therefore

2 2

14 3x y

Find the equation of the ellipse whose major and minor axes lie along the lines 3 3 0x y and 3 1 0x y

and whose lengths are 6 and 2 6 respectively.

Example – 2

Example – 3



The equation to the ellipse will obviously not be in the standard form since the axes are not along thecoordinate axes. However, we can use the coordinate axes formed by these two lines as our referenceframe

P X,Y ( )

B2

L x y2 : 3 + – 1= 0

A A1 2 = 6

A2

Y

A1 X L x – y1 : 3 + 3 = 0B B1 2 = 2 6

O

B1

Assume any point onthe ellipse asreferred to the original axes or referred to the new axes

P(x, y)

P(X, Y)

Consider an arbitrary point P on the ellipse whose coordinates are (x, y) with respect to the originalaxes (not shown) and (X, Y) with respect to the new axes, the L1 – L2 system. In this new system, theequation of the ellipse is simply

2 2

2 2 1X Ya b

where a = 3 and 6.b Thus, the equation is

2 2

19 6

X Y...(1)

Now, we wish to write the equation of the ellipse in the x-y coordinate system. For this purpose, wemake the following observations from the figure:

What is X ? It is simply the perpendicular distance of P from L2:

3 110

x yX

Similarly, Y is simply the perpendicular distance of P from L1 :

3 310

x yY

Thus, using (1), the equation of the ellipse in x-y form is

2 2(3 1) ( 3 3) 190 60

x y x y

2 221 6 29 6 58 151 0x xy y x y



As an exercise find the centre and the foci of this ellipse.

( The centre is simply the intersection of L1 and L2. The foci are at ( , 0)ae in the X -Y system.To find the foci in the x-y system, find the two points along L1 which are at a distance of ae from O oneither side of it)

Although we will not have much use for it, it is still worth mentioning that given an arbitrary fixed point P(h, k) andan arbitrary fixed straight line 0lx my n as the focus and directrix of an ellipse with eccentricity e, itsequation can be written by using the definition of an ellipse. Let (x, y) be any point on the ellipse:

Distance of ( , ) from Distance of ( , ) from the line

x y P ex y

22 2 2

2 2

( )( ) ( ) lx my nx h y k el m

This gives us the general equation of an ellipse with a given eccentricity, focus and directrix. However, we willalmost always be using the (simple !) standard form of the equation.

Given an ellipse 2 2

2 2( , ) : 1 0x yS x ya b

and a fixed point 1 1( , ),P x y how will we determine whether P lies

inside or outside the ellipse ?

The approach we will follow here is the same as the one we used in circles and parabolas to solve thesame question. Suppose P lies external to the ellipse as shown below. Drop a vertical line from Pintersecting the ellipse at Q.

yP x , y( )1 1

Q x , y( )1 0

x

Since Q lies on the ellipse, we have

2201

2 2 1yxa b

22 2 10 21 xy b

a

Example – 4



Since P lies external to the ellipse, we must have 1 0y y so that

2 21 0y y

22 2 11 21 xy b

a

2 21 12 2 1 0x y

a b

1 1( , ) 0S x y

This condition must be satisfied if P lies outside the ellipse. (Convince yourself that P maybe outside the ellipse, this condition must hold). Similarly, P lies inside the ellipse if 1 1( , ) 0.S x y Wecan write all this concisely as

1 1 1 1

1 1 1 1

1 1 1 1

( , ) 0 ( , ) lies inside the ellipse( , ) 0 ( , ) lies on the ellipse( , ) 0 ( , ) lies outside the ellipse

S x y P x yS x y P x yS x y P x y

An athlete running around a race track finds that the sum of his distances from two flag posts is always 10 metreswhile the distance between the flag posts is 8 meters. What is the area that the race track encloses ?

y

xP1 P2

AP + AP

A

1 2 is always10 metres, for any position of on the race track

A

From the situation described, the race track must be an ellipse. The eccentricity is simply

8 410 5

e

(Refer to Page -5 to understand how to find eccentricity in such a scenario).

Example – 5



If the major axis is of length 2a, we have

1 2 2 8PP ae

5a

Thus, 2(1 ) 3.b a e The equation of the elliptical race track is

2 2

125 9x y

To evaluate the area enclosed, we solve the general problem:

What is the area enclosed by the ellipse 2 2

2 2 1 ?x ya b

To find the area, we divide the ellipse into elemental strips of width dx : one such strip is shown below.

y

xx

h

x

x represents the height of an elemental strip at distance from the centre. Note from the equation of the ellipse that

h = xba a – x2 2

hx

The area of the elemental strip shown is

2 xdA h dx

2 22b a x dxa

The area of the right half of the ellipse is therefore

2 2half

0

2 abA a x dxa

2ab

The total area of the ellipse is thus

half2A A ab

For the current example, the area becomes

5 3 15 sq. mt.A



Show that the sum of the reciprocals of the squares of any two diameters of an ellipse which are at right angles toone another is a constant.

By a diameter of an ellipse, we mean any chord which passes through its centre.

Let 2 2

2 2 1x ya b

be the ellipse and let AB and CD be any two diameters of the ellipse perpendicular

to each other. y

x

CA

BD

O

Since AB and CD are diameters, we can assume AO = OB = r1, and CO = OD = r2. Also, if the slope

of AB is given by , than that of CD is obviously .2

Thus, we get the co-ordinates of A, B, C and D as

1 1 1, ( cos , sin )A B r r2 2, cos , sin

2 2C D r r

2 2( sin , cos )r r

These coordinates must satisfy the equation of the ellipse; we therefore obtain :2 2 2 2

1 12 2

cos sin 1r ra b

and 2 2 2 2

2 22 2

sin cos 1r ra b

2 2

2 2 21

1 cos sinr a b and

2 2

2 2 22

1 sin cosr a b

Adding these two relations, we have

2 2 2 21 2

1 1 1 1r r a b

2 2 2 21 2

1 1 1 1(2 ) (2 ) 4 4r r a b

2 2 2 2

1 1 1 14 4AB CD a b (a constant)

This proves the assertion stated in the question.

Example – 6



Let 2 2

2 2 1x ya b

be an ellipse. Assume a > b. A circle is described on the major axis of this ellipse as diameter.

From any point P on this circle, a perpendicular PQ is dropped onto the major axis of the ellipse. Show that PQwill always be divided in a fixed ratio by the ellipse.

y

x

P

O

R

Q

P = a , a ( cos sin )R = a , b ( cos sin )Q = a , ( cos 0)

The justification for these coordinates is given below.

As is evident, the radius of this circle, called the of the ellipse, is a, so that its equationis

2 2 2x y a

Now any point P on this circle can be taken in parametric form as ( cos , sin )P a a where isthe angle that OP makes with the horizontal.

To evaluate the y-coordinate of R, we substitute cosx a in the equation of the ellipse :

2 2 2

2 2

cos 1a ya b

siny b

Thus, R is the point ( cos , sin )a b while Q is simply ( cos ,0).a We now see that

sin sin 1sin

PR a b aRQ b b

which is independent of , proving the stated assertion.

There is one significant fact that we can learn about the ellipse 2 2

2 2 1.x ya b

We obtained the

coordinates of R as ( cos , sin ).a b This tells us that an alternative way to specify an ellipse is interms of a parameter :

cos , sinx a y b

Example – 7 AUXILIARY CIRCLE



This is referred to as the parametric form of the ellipse 2 2

2 2 1.x ya b

is called the

of the point ( , ) ( cos , sin ).x y a b It is important to note that is the angle that( cos , sin )a b makes with the horizontal; it is the angle which the corresponding point on theauxiliary circle makes with the horizontal. Depending on what value takes in the range [0, 2 )(or (- , ]), the parametric form ( cos , sin )a b gives us different points on the circumference ofthe ellipse.

The point ( cos , sin )a b in sometimes simply referred to as the point .

What are the eccentric angles of the extremities of the latus-recta in the ellipse 2 2

2 2 1 ?x ya b

As discussed earlier, the co-ordinates of the end-points of the latus-recta are 2

, .baea

Thus, if

we assume the required eccentric angle to be , we have

cosa ae

2

sin bba

This gives us four values of , given by

tan bae

corresponding to the four extremities of the two latus-recta :

y

x

A'The two latus-recta and meet the auxiliary circle in ', ', C'and '. The slopes of the linesjoining the origin to these four points give us the eccentric angles of the four extremities. Here, only one possible value of has been shown:the eccentric angle of point

AB CDA B

D

A

C'

D' B'

A

BD

C

Example – 8



Consider two points and lying on the ellipse 2 2

2 2 1.x ya b

We wish to determine the equation of the chord joining these two points.

y

x

( cos sin )a , b

( cos sin )a , b

Using the two point form, we have

sin sin sincos cos cos

y b b bx a a a

cossin 2cos sin

2

y b bx a a

cos sin cos cos sin sin2 2 2 2

x ya b

cos sin cos2 2 2

x ya b

This is the most general equation of a chord joining any two arbitrary points and on the ellipse. As an exercise using this form try writing

(a) the equation of any chord passing through the origin

and (b) the equation of the latus-recta by using the eccentric angles of its extremities which we derived earlier.

Suppose that the chord joining the points 1 and 2 on the ellipse 2 2

2 2 1x ya b

intersects the major-axis in (h, 0).

Show that1 2tan tan .

2 2h ah a

Example – 9



By the result we just derived, the equation of the chord joining 1 and 2 is

1 2 1 2 1 2cos sin cos2 2 2

x ya b

If this passes through (h, 0), we have

1 2 1 2cos cos2 2

ha

1 2

1 2

cos2

cos2

ha

1 2 1 2

1 2 1 2

cos cos2 2

cos cos2 2

h ah a

1 2tan tan2 2

A circle intersects the ellipse 2 2

2 2 1x ya b

in four points A, B, C and D whose eccentric angles are respectively

1 2 3, , and 4. Show that 1 2 3 4 will be an integral multiple of 2 .

Suppose that the circle cuts the ellipse as shown :

y

x

A( )1

B( )2

C( )3D( )4

Example – 10



Using the general equation of a chord joining two arbitrary points on an ellipse derived earlier, we canwrite 1 0L and 2 0,L the equation of AB and CD respectively. Doing this has the advantage thatwe can now write (using a family of circles approach) any circle passing through the four point A, B,C and D as

1 2 0S L L

where S = 0 is the equation of the ellipse. Imposing the necessary condition of for this equation torepresent a circle will finally yield the constraint we are actually looking for.

Thus, any curve through A, B, C, D has the form

3 4 3 41 2 1 22 2

2 21 2 3 4

Equation of Equation of

cos sincos sin2 22 2

1 0cos cos

2 2AB CD

x yx ya ba bx y

a b

This represents a circle if

2 2Coeff. of Coeff. of x y

Coeff. of 0xy

3 4 3 41 2 1 22 2 2 2

1 1cos cos sin sin2 2 2 2a a b b

and 3 4 3 41 2 1 2cos sin cos sin 02 2 2 2

The second relation gives

1 2 3 4cos 02

1 2 3 4

2n , n

1 2 3 4 2n , n

This is the desired result.

The result we’ve derived in this question is quite significant and has a good amount of use; it would therefore beadvantageous to remember it.



If 0px qy r cuts the ellipse 2 2

2 2 1x ya b

in points whose eccentric angles differ by ,2 show that

2 2 2 2 22a p b q r

The two points of intersection can be assumed to be

( cos , sin )P a b

cos , sin2 2

Q a b

( sin , cos )a bSince P and Q both lie on the line 0,px qy r their coordinates satisfy its equation. We thus have

cos sin 0ap bq rand sin cos 0ap bq r

cos sinap bq r and sin cos .ap bq rSquaring and adding the two relations gives us the described result.

Let P be a point on the ellipse 2 2

2 2 1, 0 .x y b aa b

Let the line parallel to the y-axis passing through P meet

the circle 2 2 2x y a at the point Q such that P and Q are on the same side of the x-axis. For two positive real

numbers r and s, find the locus of the point R on PQ such : :PR RQ r s as P varies over the ellipse.

The circle 2 2 2x y a is the auxiliary circle of the given ellipse.y

x

Q a , a ( cos sin )

P a , b ( cos sin )

R

Example – 11

Example – 12



We can assume point P to be ( cos , sin )a b so that Q will be ( cos , sin ).a a The point R(h, k)divides the segment PQ internally in the ratio r : s. Thus,

cos cos sin sin,ar as ar bsh kr s r s

We need to eliminate from these two relations to obtain a relation between h and k. Thus,

( )cos , sinh k r sa ar bs

Squaring and adding the two gives us the required relation :2 2 2

2 2

( ) 1( )

h r s ka ar bs

The required locus of R is2 2 2

2 2

( ) 1( )

x r s ya ar bs

which is an ellipse, as might have been expected.

Consider the ellipse 2 2

2 2 1.x ya b

Let F be its focus (ae, 0) and S be its vertex (a, 0). Consider any point P on the

ellipse whose eccentric angle is , while .SFP Prove that

1tan tan2 1 2

ee

y

x

P a , b ( cos sin )

T F ae,( 0) S a,( 0)

Drop a perpendicular from onto the major

axis. The point is then ( cos 0)

PT PT

T = a ,

We have, from the figure,

cos FTPF

Example – 13



Since F is the focus, note that PF will simply be e times the distance of P from the directrix .axe

Thus,

cosaPF e ae

cosa ae

(cos ) coscos(1 cos ) 1 cos

a e ea e e ...(1)

Thus,

2 1 cos (1 )(1 cos )tan2 1 cos (1 )(1 cos )

ee {Using (1)}

2(1 ) tan

(1 ) 2ee

1tan tan2 1 2

ee

Consider a focal chord AB of the ellipse 2 2

2 2 1x ya b

where the eccentric angles of A and B are 1 and 2

respectively. If e is the eccentricity of the ellipse, prove that

1 2

1 2

sin sinsin( )

e

The chord joining A and B has the equation

1 2 1 2 1 2cos sin cos2 2 2

x ya a ...(1)

Since AB is a focal chord of the ellipse (say, it passes through 1( ,0)),F ae the coordinates of the focusmust satisfy (1) so that we have

1 2 1 2cos cos2 2

e

1 2

1 2

cos2

cos2

e...(2)

Example – 14



Multiplying the numerator and denominator of the RHS of (2) by 1 22sin ,2

we have

1 2 1 2

1 2 1 2

2cos sin2 2

2cos sin2 2

e

1 2

1 2

sin sinsin( )

Consider an ellipse 2 2

2 2 1x ya b

and a variable line .y mx c What is the condition on m and c such that the

line

intersects the ellipse in two distinct points ?

touches the ellipse ?

does not intersect with the ellipse ?

As we’ve done in the case of circles and parabolas, to find the intersection (points) of the line and theellipse, we must solve their equations simultaneously;

2 2

2 2 1 ;x y y mx ca b

2 2

2 2

( ) 1x mx ca b

2 2 2 2 2 2 2 2( ) 2 ( ) 0a m b x a mcx a c b ...(1)

The line y mx c

intersects the ellipse

touches the ellipse

does not touch / intersect the ellipse

accordingly as the quadratic (1) has its discriminant greater than , equal to or less than 0.

The condition for tangency (D = 0) is of special intersect. Verify that it comes out to be2 2 2 2c a m b

Thus, we can say that the line 2 2 2y mx a m b will always be a tangent to the ellipse, whatevermay be the value of m. We discuss tangents in more detail in the next section.

Example – 15 INTERSECTION OF A LINE WITH AN ELLIPSE



Let P be a variable point on the ellipse 2 2

2 2 1x ya b

with foci S1 and S2. Find 1 2max(area( )).PS S

Find the equation of the ellipse with foci at (0, 4) and eccentricity 4 .5

Show that 2 24 2 16 13 0x y x y is the equation of an ellipse. Where are its foci ?

Find the equation of the ellipse whose foci are ( 2,3) and whose semi-minor axis is of length 5.

A straight rod of length l slides between the x-axis and the y-axis, as shown. Show that the locus ofits mid-point is an ellipse. What is its eccentricity ?

y

x

l

Show that the traingle with vertices (1, 2), (3, –1) and (–2, 1)lies completely inside the ellipse2 22 13.x y

TRY YOURSELF - I

Ellipse-1 Basics

Documents

Transcript of Ellipse-1 Basics