Mathematics

EllipseSession - 2

Session Objectives

Session Objectives

1. Equation of the tangent in point (x1, y1) form

2. Equation of normal in point (x1, y1) form

3. Equation of tangent and normal in parametric form

4. Number of tangents drawn from a point to an ellipse

5. Director circle

6. Equation of pair of tangents

7. Equation of chord of contact

8. Equation of normal in slope (m) form

9. Number of normals (co-normal points)

10. Equation of chord whose middle point is given

11. Diameter of ellipse

12. Conjugate diameters

Equation of the Tangent in Point(x1, y1) Form

Equation of tangent to ellipse

at (x1, y1) is given by

2 2

2 2

x y+ = 1

a b

1 12 2

xx yy+ = 1 T = 0

a b

2

11 12

1

b xSlope of tangent =– , point of contact = x , y

a y

Equation of the Tangent in Point(x1, y1) Form

Working rule for finding T = 0,

replace

and keeping constant unchanged.

2 21 1 1x by xx , y by yy , 2x by x + x

1 1 12y by y + y , 2xy by xy + x y

Equation of Normal in Point (x1, y1) Form

Equation of normal at to ellipse

given by

1 1x , y

2 2

2 2

x y+ = 1 is

a b2 2

2 2 1 12 2

1 1 1 1

x – x y – ya x b y– = a – b or =

x y x /a y /b

2

12

1

a ySlope of normal =

b x

Equation of Tangent and Normal in Parametric () Form

Equation of tangent at

using point form is

2 2

2 2

x ya cosθ, bsinθ to + = 1

a b

2 2

x a cos ybsin1 or

a b

x ycosθ + sinθ = 1

a b

Slope point of contact

with the ellipse.

–b= cotθ,

a a cosθ, b sinθ

Equation of Tangent and Normal in Parametric () Form

Equation of normal at

becomes

2 2

2 2

x yacos , bsin to 1

a b

2 22 2a x b y

a bacos b sin

2 2ax byor – a – b

cos sin 2 2or a x sec bycosec = a b

Slope = foot of normal atanθ,

b a cosθ, bsinθ

Number of Tangents Drawn From a Point to an Ellipse

Two tangents can be drawn from a point toan ellipse it may be real or imaginary.

y

xO

Number of Tangents Drawn From a Point to an Ellipse

2 2 2 2 2Observation from m h - a - 2hkm + k - b = 0

2 22 2 2 2 2 2 2 2

2 2

h kD 4h k 4 h a k b 4a b 1

a b

(i) Real and distinct if , i.e. point liesoutside the ellipse.

2 2

2 2

h kD 0 1

a b

(ii) Real and coincident if , i.e. pointlies on the ellipse.

2 2

2 2

h kD 0 1

a b

(iii) Imaginary if , i.e. point lies insidethe ellipse.

2 2

2 2

h kD 0 1

a b

(iv)

2 2

1 2 1 22 2 2 2

2hk k bm + m = , m m =

h a h a

Director Circle

Director circle is the locus of point of intersection of perpendicular tangents to the conic.

Equation of director circle of ellipse

Let (h, k) be the point of intersection of tangents to

and slope of tangent be m. Then we have 2 2

2 2

x y1

a b

2 2 2 2 2m h a 2hkm k b 0

Director Circle

2 2

1 2 2 2

k bm m

h a if tangents are

perpendicular, then , i.e.

2 2

2 2

k b1

h a

2 2 2 2h k a b

Hence, locus of (h, k) is .2 2 2 2x + y = a +b

Equation of Pair of Tangents

As we have seen earlier from a point(h, k) lying outside the ellipse, wehave two real and distinct tangentspossible. Combined equation ofthese tangents is given by

22 2 2 2

2 2 2 2 2 2

x y h k hx ky1 1 1

a b a b a b

or haveusual meanings.

21 1SS T , where S, S ,T

y

xO

(h, k)

Note: We can obtain above equation by eliminating ‘m’ from

y – k = m(x – h) and

2 2 2 2 2m h a 2hkm k b 0

Equation of Chord of Contact

Equation of chord of contact of point

(h, k) outside the ellipse is 2 2

2 2

x y1

a b

2 2

hx ky+ = 1

a bor T = 0

y

xO

(h, k)

Q

P ( )

PQ is chord of contact of point (h, k).

Equation of Normal Slope (m) Form

Equation of normal of slope m to is

2 2

2 2

x y+ = 1

a b

2 2

2 2 2

a by = mx m

a +b m

Slope = m, foot of normal is acosθ, bsinθ

Number of Normals (Co-normal Points)

Four normals can be drawn from a pointto an ellipse.

Equation of Chord Whose Middle Pointis Given

Equation of chord of

bisected at (h, k) is ,

i.e

2 2

2 2

x y+ = 1

a b2

2 2 2 2

hx ky h ky+ – 1 = + – 1

a b a b

1T = S

where T, S1 have usual meanings.

Diameter of Ellipse

The locus of mid-point of a system of parallel chords of an ellipse is called diameter and chords are called its double ordinates. The end points of the diameter lying on the ellipse are called vertices of diameter.

Equation of diameter of ellipse

Let the system of parallel chords be given by y = mx + c, where ‘m’ is fixed and ‘c’ is a variable. Let (h, k) be its middle point. Then equation of chord with middle point at (h, k) is given by

2 2

1 2 2 2 2

hx ky h kT S , i.e. – 1 – 1

a b a b

Diameter of Ellipse

Then its slope is m.

Hence, 2

2

–b hm

a k

Locus of (h, k) is or

2

2

b xm or

a y

2

2

by = x

a m

which is the required equation of diameter. Note thatdiameter passes through the centre of ellipse.Hence, equation of diameter bisecting the parallel

chords of slope ‘m’ of ellipse

2 2 2

2 2 2

x y b1 is y x.

a b a m

Conjugate diameters

Two diameters of an ellipse are said tobe conjugate diameters. If eachbisects the chords parallel to the other.

Condition of conjugate diameters

Let be two conjugate diameters

of (Recall that diameter of ellipse passes

through the centre of the ellipse.). Then diameter

bisecting the chords parallel to is given by

1 2y m x and y m x

2 2

2 2

x y1

a b

1y m x

2

21

by x

a mwhich is given as y= m2x.

Conjugate diameters

Then

2

2 21

bm

a mor

2

1 2 2

bm m =

a

Thus, are conjugate diameters

of ellipse .

1 2y m x and y m x

2 2 2

1 22 2 2

x y b1 if m m

a b a

Note: Major axis and minor axis are conjugatediameters, as each bisects the chords parallelto the other but product of their slopes is notdefined.

Properties of conjugate diameters

(i) The eccentric angles of the ends of apair of conjugate diameters of an ellipse

differ by

.2

y

xO

y = m x1y = m x2

QS

PR(a cos , b sin ) (a cos , b sin )

Properties of conjugate diameters

(ii)The sum of the squares of any twoconjugate semi-diameters (half of thediameter) is constant and is given bysum of squares of semi-axis, i.e.

2 2 2 2OP OR cons tant a b

Note: That major axis and minor axis are also conjugate diameters.

(iii)The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter passing through this point.

Properties of conjugate diameters

y

xO

QS

P ( )R

+ –2

A

B

C

D

+ —32

+

(iv) Tangents at the ends of the pair ofconjugate diameters form aparallelogram, i.e. ABCD is aparallelogram.

Properties of conjugate diameters

(v) The area of the parallelogramformed by the tangents at the endsof conjugate diameters is constantand is given by the product of theaxes, i.e. area (ABCD) = 4ab.

y

xO

QS

P ( )R

+ –2

A

B

C

D

+ —32+

Concyclic Points

Any circle intersects an ellipse in two or four real points. They are called concyclic points.

If , , , be the eccentric angles of four concyclic points on an ellipse, then

+ + + = 2n, i.e. even multiple of .

P( )Q( )

R( ) S( )

Class Exercise - 1

The area of the quadrilateral formed bythe tangents at the end points of latus

rectum to the ellipse is 2 2x y

19 5

(a) sq. units (b) 9 sq. units

(c) 27 sq. units (d) sq. units

27

4

27

2

Solutiony

xO

A

B

C

D

(– 3, 0)( 3, 0)

– 2, –53

(0, 5)2, –5

3

2, – 5 3

–

– 2, – 5 3

–

(0, 5)–

End points of latus rectum are given by

2b 5ae, 2, .

a 3

Solution contd..

Tangents at these points, using point

form, are given by 2x 5 y

19 3 5

or 2x 3y 9

Equation of AD = 2x + 3y = 9 9

OA 3, OD2

Area of the parallelogram ABCD = 4 × Area AOD

1

4. .AO.OD2

9

2.3. 27 sq. units2

Hence, answer is (c).

Class Exercise - 2

Find the point on which isnearest to the line x + y = 7.

2 2x 2y 6,

Solutiony

xO

P

3

P

6

Q

7

7

6–3–

The point which is nearest to the line is the point in the first quadrant, where tangent is parallel to PQ or if PQ is moved parallel to itself towards ellipse, the point where PQ first meets, i.e. touches the ellipse is the required point.

Solution contd..

Let be the required point,

tangent at is

1 1x , y

1 1x , y

1 1xx 2yy 6 its slope 1

1

1 11

x1 x 2y

2y

Also lies on 1 1x , y 2 2x 2y 6

2 2 2 21 1 1 1 1 1x 2y 6 4y 2y 6 y 1 x 2

(2, 1) is the required point.

Note: (–2, –1) is the farthest from the line x + y = 7

Solution contd..

Alternative:

Slope of tangent is –1.

Point of contact is

2 2

2 2 2 2 2 2

2 2

2 2 2 2 2 2

a m b,

a m b a m b

a m bor – , –

a m b a m b

6 1 3,

6 3 6 3

2, 1

Required point is (2, 1).

Class Exercise - 3

If the normal at the end of a latus rectumof an ellipse of eccentricity ‘e’ passesthrough one end of the minor axis, then

2 4e e

(a) 1 (b) 2

(c) 3 (d) 4

Solutiony

xO

ae, –ba

2

(0, –b)

Slope of normal at . 2

11 1 2

1

a yx , y is

b x

Slope of normal at .

2 2

2

b ab 1ae, is or

a eb ae

Solution contd..

Equation of normal at is

2bae,

a

2b e

x ey aea

If it passes through (0, –b), then

be = ae – or ab = a2 – b2 = a2 e2

2b ea

2 2 4 4a b a e

2 2 2 4 4 4 2a a 1 e a e e e 1

Solution contd..

Short cut:

As e < 1 for ellipse

2 4e e 2

Now check with options.

is the only possibility. 2 4e e 1

Hence, answer is (a).

Class Exercise - 4

A tangent to the ellipse

meets the ellipse

in the points P and Q. Prove that the tangents at P and Q are at right angles.

2 2

2 2

x y1

a b

2 2x y

a ba b

Solution

Let the tangents at P, Q intersect atR(h, k). Then according to the givenconditions, chord of contact of

R w.r.t. touches

.

2 2x y

a ba b

2 2

2 2

x y1

a b

i.e. 2 2

2 2

hx ky x ya b touches 1

a b a b

bh by x a b

ak kis tangent to

2 2

2 2

x y1

a b

Solution contd..

22

2 2 2 2 2 2 22

b bha b a b c a m b

akk

2 2 2 2 2 2a b b b h b k 2 2 2or a b h k

Hence, locus of (h, k) is

22 2x y a b a a b b a b

which is the equation of director circle of

2 2x y1

a a b b a b

Hence, tangents at P, Q intersect at right angle.

Class Exercise - 5

Find the locus of mid-points of normal

chords of the ellipse 2 2

2 2

x y1.

a b

Solution

Let (h, k) be the mid-point. Then its

equation is given by . 2 2

2 2 2 2

hx ky h k

a b a b

If it is normal at , then this equation is same as ' ' 2 2ax sec bycosec a b

2 2

2 2

2 2 2 2

asec bcosec a bh k h ka b a b

Solution contd...

2 2 2 2

2 2 2 2

2 2 2 22 2

h k h ka b

a b a bcos , sin

h ka b a b

a b

2 2As sin cos 1

22 2

2 2 6 6

2 2 22 2

h k

a b a b1

h ka b

Locus of (h, k) is

22 2 6 6 22 22 2 2 2

x y a ba b

a b x y

Class Exercise - 6

If P and D be the ends of semi-conjugatediameters, find the locus of foot ofperpendicular from centre upon PD.

Solution

P acos , bsin ,

D acos , bsin2 2

asin , bcos

Equation of PD is

x y

cos sin cos ...(i)a 4 b 4 4

[Using equation of chord joining ] and

Equation of line perpendicular to aboveline passing through (0, 0) is given by

x ysin cos 0

b 4 a 4

Solution contd..

bysin cos

4 ax 4

From (i) we get

2 2 2

cos axcos4 4cos

4 x y x y

a ax

2 2

bycos4sin

4 x y

22 2 2 2 2 2 2 2sin cos 1 a x b y 2 x y4 4

Class Exercise - 7

Find the locus of the middle points of chord of an ellipse which are drawn through the positive end of the minor axis.

Solution

Let (a cos , b sin ) be the coordinate ofthe other extremities of the chord of

ellipse . The positive end of the

minor axis is clearly (0, b).

2 2

2 2

x y1

a b

Let (x, y) be the mid-point of the chord.

a cos 0x

2

b bsiny ...(i)

2

2x 2y b

cos , sin ...(ii)a b

Solution contd..

Squaring and adding (i) and (ii),

22 2 2 2

2 2 2

4x 2y b 4x 4y 4yb b

ba a b

On simplifying the required locus of

(x, y) is

2 2

2 2

x y y

ba b

Class Exercise - 8

Prove that the area of the triangle form by three points on an ellipse, whose eccentric angles are is

Prove also that its area is to the area of the triangle formed by the corresponding points on the auxiliary circle as b : a.

, , and

2absin sin sin

2 2 2

Solution

Let the coordinates of the given points

on the ellipse be

be

2 2

2 2

x y1

a b acos , bsin ,

acos , bsin acos , bsin

Area of the triangle formed by these points

1 acos bsin1

1 acos bsin2

1 acos bsin

1

ab sin sin sin2

Solution contd..

Using

C D C – D

sinC sinD 2sin cos , etc.2 2

2absin sin sin ... (i)2 2 2

Second Part:

Clearly, coordinates of the points of the auxiliary circle corresponding to the given three points may be obtained by putting b = a.

2So area of 2a sin sin sin ... (ii)

2 2 2

bThe ratio of the area is .

a

Class Exercise - 9

Prove that the straight line

is a normal to the ellipse if

.

x my n

22 22 2

2 2 2

a ba b

m nl

Solution

The normal at any point

of the ellipse is given by

P acos ,bsin

2 2

2 2x y

1a b

2 2ax bya b

cos sin

Comparing the given line, and the equation of the normal

x my n

2 2ax bya b

cos sin

2 2

cos msin na b a b

Solution contd..

2 2 2 2

an bncos , sin

a b m a bl

Squaring and adding

2 2

2 22 2 2 2

an bncos sin

a b m a b

22 22 2

2 2 2

a ba bi.e.

m n

Class Exercise - 10

The tangents drawn from a point P to the ellipse make angles 1 and 2 with the major axis. Find the locus of P when 1 + 2 is constant = 2.

Solution

Let the equation of the tangent to the

ellipse,

and let P(h, k) lies on the tangent.

2 2

2 2 22 2

x y1 be y mx a m b

a b

Then 2 2 2k mh a m b

2 2 2 2k mh a m b

2 2 2 2 2m h a 2mhk k b 0

1 1 2 2Let tan m and tan m

Solution contd..

1 2 1 22 2

2hkm m tan tan

h a

2 22 2

1 2 1 22 2 2 2

k bk bm m or tan tan

h a h a

1 2By hypothesis 2 .

1 2tan tan2 = Constant.

1 2

1 2

tan tantan2

1 tan tan

Solution contd..

2 2

2 2

2 2

2hk

h atan2

k b1

h a

2 2 2 2Locus is x 2xycot 2 y a b

Thank you