Basic Concepts of Ellipse

7/30/2019 Basic Concepts of Ellipse

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Basic ConceptsIf we slice an egg obliquely there will appear a typical curve by its edge. We find thesimilar but larger curve if we trace the curve of the movement of earth around the sun.

Our mathematicians and scientist, named this curve as the ellipse. Ellipse one of theconic sections is obtained by cutting one nappe of cone with a plane that does not passthrough the vertex.

Definition

An ellipse is locus of a point, which moves in a plane such that the ratio of its distancesfrom a fixed point and a fixed line is constant and always less than one.

In other words "Ellipse" is a conic for which the eccentricitye < 1. Let S be the focus ofellipse, P any point on the ellipse and PM perpendiculardistance of the directrix from P, then

SP/PM = e < 1

Let S be the focus and ZM be the directrix of the ellipse. Let be itseccentricity.

We draw SZ perpendicular to the directrix and divide SZ internally and externally in theratio e : 1 and let A and A' be the internal and external point of division.

Then we have SA = e AZ ...... (1)

And SA' = e A'Z ...... (2)

.. A and A' lie on the ellipse.

Let AA' = 2a and take O the midpoint of AA' as origin. Let P(x, y) be any point ontheellipse referred to OA and OB as co-ordinate axis.

Then from figure it is evident that

AS = AO - OS = a - OS

AZ = OZ - OA = OZ - a


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A'S = A'O + OS = a + OS

A'Z = OZ + OA' = OZ + a

Substituting these values in (1) and (2), we have

a - OS = e (OZ - a) ...... (3)

a + OS = e (OZ + a) ...... (4)

Adding (3) and (4), we get

2a = 2 e OZ

Or OZ = a/e ...... (5)

Subtracting (3) from (4), we get

2 OS = 2ae => OS = ae ...... (6)

.. The directrix MZ is x = OZ = a/e and the co-ordinate of the focus S are(OS, 0) i.e. (ae, 0). Now as P(x, y) lies on the ellipse.

So we get

SP = e PM or SP2 = e2 PM2

(x - ae)2 + y2 = e2 [OZ - x co-ordinate of P]2

=> (x - ae)2 + y2 = e2 [a/e - x]2 = (a - ex)2 ...... (7)

=> x2 + a2e2 - 2axe + y2 = a2 + e2x2 - 2aex

or x2/a2 + y2/a2(1-e2) = 1 [Dividing each term by a2 (1 - e2)]

or x2/a2 + y2/b2 = 1 where b2 = a2 (1 - e2)

This is the standard equation of an ellipse, O is called the centre of the ellipse, AA'and BB' are called the major and minor axes ofellipse (where b < a).

There exists a second focus and second directrix for the curve. On the negative sideof the origin take a point S', which is such that SO = S'O = ae and another point Z' suchthat ZO = OZ' = a/e.

Draw Z'K' perpendicular to ZZ' and PM' perpendicular to Z'K'

The equation (7) may also be written in the form

(x + ae)2 + y2 = (a + ex)2

=> S'P2 = e2 (PM')2


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Hence, any point P on the curve is such that its distance from S' is e times to itsdistance from Z'K' so we should have obtained the same curve, if we had started with S'as focus, a Z'K' as directrix and the same eccentricity.

Pause:

We have considered a > b, now if we consider b > a, what will be the shape ofthe ellipse x2/a2 +y2/b2? In this case the major axis AA' of the ellipse is along the y-axis and is of length 2b. See figure.

The minor axis of BB' = 2a. The foci S and S' are (0, be) and (0, -be) respectively.The directrix are MZ and M'Z' given by y = + b/e, respectively. Also here a 2 = b2 (1 -

e2

).

Note:

Let P(x1, y1) be any point. This point lies outside, on or inside the ellipse (8) accordingas x21/a

2 + y21/b2 = 1 > 0 or = 0 or < 0.

Central Curve

A curve is said to be a central curve if there is a point, called the centre, such that everychord passing through it is bisected at it.

Latus rectum:

The length of a chord through the focus and at right angle to the major axis ofthe ellipse is known as the latus rectum of the ellipse.


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There being two foci of an ellipse, there are two rectum, which are of equallength.

yL = b2/a

.. The length of latus rectum LSL' = 2b2/a

Notes:

The major axis AA' is of length 2a and the minor axis BB' is of length 2b. The foci are (-ae, 0) and (ae, 0). The equations of the directrices are x = a/e and x = -a/e. The length of the semi latus rectum = b2/a. Circle is a particular case of an ellipse with e = 0.

Focal Distance of a Point

Since S'P = ePN', SP = ePN,

S'P + SP = e(PN + PN')

= e (NN') = e(2a/e) = 2a

=> the sum of the focal distances of any point on the ellipse is equal to its majoraxis.

Another definition of an ellipse


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Let ellipse be x2/a2 +y2/b2 = 1 ...... (i)

Its foci S and S' are (ae, 0) and (-ae, 0). The equation of its directrices MZand M'Z' are x = a/e and x = -a/e respectively. Let P(x1, y1) be any point on (i)

Now SP = e PM = e NZ = e (OZ - ON) = e[(a/e)-x1] = a - ex

1

and S'P = ePM' = e (Z'N) = e (OZ' + ON) = e[(a/e) + x1] = a + ex1

.. SP + S'P = 2a = AA'

So by this property an ellipse can also be defined as "the locus of a point which movessuch that the sum of its distances from two fixed point is always constant.

Other Forms

If in the equation x2/a2 +y2/b2 = 1, a2 < b2, then the major and minor axis oftheellipse lie along the y and the x-axis and are of lengths 2b and 2a respectively. Thefoci become (0, + be), and the directrices become y = + b/e where e = (1-a2/b2 ). Thelength of the semi-lactus rectum becomes a2/b2.

If the centre of the ellipse be taken at (h, k) and axes parallel to the x and the y-axes,then the equation of the ellipse is (x-h)2/a2 +(y-k)2/b2 = 1.

Let eh equation of the directrix of an ellipse be ax + by + c = 0 and the focus be (h,k).

Let the eccentricity of the ellipse be e(e < 1).

If P(x, y) is any point on the ellipse, then

PS2 = e2 PM2


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=> (x - h)2 + (y - k)2 = e2(ax+by+c)2/(a2 + b2 ), , which is of the form

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ... (*) where

= abc + 2fgh + af2 - bg2 - ch2 0, h2 < ab.

These are the necessary and sufficient conditions for a general quadratic equation givenby (*) to represent an ellipse.

Position of a Point Relative to an Ellipse

The point P(x1, y1) is outside or inside the ellipse x2/a2 + y2/b2 = 1, according as the

quantity ((x12

)/a2

+(y12

)/b2

-1) is positive or negative.

Parametric Equation of an EllipseClearly, x = a cos, y = b sin satisfy the equation x2/a2 +y2/b2 = 1 for all real valuesof .

Hence, the parametric equations of the ellipse x2/a2 +y2/b2= 1 are x = a cos,y = b sinq where is the parameter.

Also (a cos , b sin ) is a point on theellipse x2

/a

2

+y

2

/b

2

= 1 for all values of(0 < < 2).

The point (a cos, b sin) is also called the point . The angle is called the eccentricangle of the point (a cos, b sin) on theellipse.

Draw a circle with AA' (the major axis) as the diameter. This circle is called the auxiliarycircle of the ellipse. The equation of the circle is x2 + y2 = a2. Any point Q on the circleis (a cos, a sin). Draw QM as perpendicular to AA' cutting the ellipseat P. The x-co-ordinate of P = CM = a cos.

=> y-co-ordinate of P is b sin

=> P (a cos, b sin).


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Illustration:

Find the centre, the length of the axes and the eccentricity of the ellipse 2x2 + 3y2 -

4x - 12y + 13 = 0.

Solution:

The given equation can be written as 2(x - 1)2 + 3(y - 2)2 = 1

=>(x-1)2/(1/2)+(y-2)2/(1/3) = 1 => The centre of the ellipse is (1, 2).

The major axis = 2. 12 = 2.

The minor axis = 2.1/3=2/3 => e2 = 1 1/2 = 1/3 => e = 1/3.

Illustration:

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

Solution:

Here S is (2, 3), S' is (-2, 3) and b = 5.

=> SS' = 4 = 2ae ae = 2.

But b2 = a2 (1 - e2) => 5 = a2 - 4 => a = 3.

Centre C of the ellipse is (0, 3).

Hence the equation of the ellipse is (x-0)2/9+(y-3)2/5 = 1

=> 5x2 + 9y2 - 54y + 36 = 0.

Illustration:


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Find the equation of theellipse having centre at (1, 2), one focus at (6, 2) andpassing through the point (4, 6).

Solution:

With centre at (1, 2) the equation of the ellipse is (x-1)2/a2 +(y-2)2/b2 = 1. It passesthrough the point (4, 6).

=> 9/a2 +16/b2 = 1. ...... (1)

Distance between the focus and the centre = (6 - 1) = 5 = ae

=> b2 = a2 - a2e2 = a2 - 25. ...... (2)

Solving for a2 and b2 from the equations (1) and (2), we get a2 = 45 and b2 = 20.

Hence the equation of the ellipse is (x-1)2/45+(y-2)2/20 = 1.

Illustration:

Find the equation of theellipse (in standard form) having latus rectum 5andeccentricity 2/3.

Solution:

Let the ellipse be x2/a2 +y2/b2 = 1 with a > b.

Latus rectum = 5 = 2b2/a => 2b2 = 5a. ...... (1)

Also b2 = a2 (1 - e2) = a2(1-4/9) = 5a2/9

=> 5a/2 = 5a2/9 => a = 9/2 and hence b2 = 5/2a = 45/4.

The equation of theellipse, in the standard form, is thus x2/(81/4)+y2/(45/4) = 1.

Illustration:

Find the equation of the ellipse, which cuts the intercept of length 3 and 2 on

positive x and y-axis. Centre of the ellipse is origin and major and minor axes are alongthe positive x-axis and along positive y-axis.

Solution:

x2/a2 +y2/b2 = 1 ...... (1)

According to the given condition the ellipse (1) passes through (3, 0) and (0,2), so we have.

9/a2 = 1 => a2 = 9

And 4/b2 = 1 => b2 = 4


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Therefore, the equation of the ellipse is x2/9 + y2/4 = 1

Illustration:

Obtain the equation of an ellipse whose focus is the point (-1, 1) whose directrix isthe line passing through (2, 5) having the unit gradient and whose eccentricity is .

Solution:

Let P(x, y) be any point on ellipse.

Its focus is S (-1, 1).

Let the directrix be y = x + c ...... (1)

(. gradient m = 1)

Line (1) passes through (2, 5) so,

5 = 2 + c => c = 3

The directrix is y = x + 3

=> x - y + 3 = 0 ...... (2)

Now let PM be the perpendicular from P, drawn to its directrix

(2). By definition of ellipse SP = e PM

or SP2 = e2 PM2

=> (x + 1)2 + (y - 1)2 = (1/2)2 [(x-y+3)/((12+12 ))]2

=> 8[(x + 1)2 + (y - 1)2] = (x - y + 3)2

=> 7x2 + 7y2 + 2xy + 10x - 10y + 7 = 0,

This is the required equation of ellipse.

Tangent and NormalThe ellipse is x2/a2 +y2/b2 = 1 ...... (1)

Let P(x1, y1) and Q(x2, y2) be two points on the ellipse. The equation of the line PQ is,

y - y1 = ((y2-y1)/(x2-x1 )) (x - x1) ...... (2)

since points P and Q lies on (1), we get

(y2-y1)/(x2-x1 )=(-b2 (x2-x1 ))/(a

2 (y2-y1 ) )


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So (2) becomes

y - y1 =(-b2 (x2-x1 ))/(a

2 (y2-y1)) (x - x1)

As point Q approaches towards point P along the ellipse, the line PQ tends tothetangent at P. So, by substituting x1 and y1 for x2 and y2 in the above equation, wehave the equation of the tangent at P as

y - y1 = (-b2 (2x1 ))/(a

2 (2y1)) (x - x1)

=> (xx1)/a2 + yy1/b

2 =(x12)/a2 +(y`

1)/b2 = 1 [as P lies on (1)]

Hence the equation of the tangent to x2/a2+y2/b2= 1 at P(x1,y1) is xx1/a2 + yy1/b

2=1. ...... (3)

Equation of the tangent in terms of 'm'

Let the line y = mx + c ....... (4)

Touch the ellipse x2/a2 +y2/b2 = 1 ...... (5)

Eliminating y between (4) and (5), we get;

x2 (b2 + a2m2) + 2a2mcx + a2(c2 - b2) = 0 ...... (6)

If (4) touches (5) then the roots f (6) must be coincident i.e. D = 0

i.e. (2a2mc)2 = 4(b2 + a2m2) a2(c2 - b2)

solving this we get c = +(a2 m2+b2)

So the equation oftangent is y = mx + (a2 m2+b2) for all real m ......(7)

From the equation (6) and (7) we get the point of contact as ((+_a2m)/(a2

m2+b2 ),(b2)/(a2 m2+b ))

Tangent and Normal

Equation of the Tangent to an Ellipse

* Let the equation of the ellipse be x2/a2 +y2/b2 = 1 . Slope oftangent at

(x1, y1) = dy/dx(x1,y1)=-b2/a2 x2/y1

Hence the equation of the tangent at (x1, y1) is y - y1 = (-b2 x1)/(a

2 y1)(x - x1)

=> (xx1)/a2 +(yy1)/b

2 = 1.


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* Equation oftangentat the point q i.e. (a cos, b sin) is obtained by putting x1= acos, y1= b sin

=> (x cos )/a+(y sin )/b = 1.

Equation of the Tangent in Terms of its slope; Condition of Tangency

To find the condition that the line y=mx+ c may touch the ellipse x2/a2 +y2/b2 = 1.

Tangent to the ellipseat (a cos, b sin) is (x cos)/a+(y sin)/b = 1.

If y - mx = c is also a tangent to the given ellipse at 'q', then comparing coefficients,we get

(cos)/am = (sin)/b=1/c or 1/c =(sin)/b=(cos)/(-am)=((sin2+cos2)/(b2+(-

am)2))=1/(a2 m2+b2 )

=> c = (a2 m2+b2 ).

Therefore, the equation of a tangent to the ellipse x2/a2 +y2/b2 = 1 is

y = mx (a2 m2+b2 ) for all values of m.

Illustration:

Find the locus of the point of intersection of the tangents tothe ellipse x2/a2+y2/b2 = 1 (a > b) which meet at right angles.

Solution:

The line y = mx (a2 m2+b2) is a tangent to the given ellipse for all m. Let is passesthrough (h, k).

=> k - mh = (a2 m2+b2 ) => k2 + m2h2 - 2hkm = a2m2 + b2

=> m2 (h2 - a2) - 2hkm + k2 - b2 = 0.

If the tangents are at right angles, then m1m2 = -1.

=> (k2-b2)/(h2-a2 ) = - 1 => h2 + k2 = a2 + b2.

Hence the locus of the point (h, k) is x2 + y2 = a2 + b2 which is a circle. This circle iscalled the Director Circle of the ellipse.

Tangent and Normal

Note:


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The locus of the point of the intersection of two perpendicular tangents to anellipse is acircle known as the director circle.

Illustration:

Prove that the locus of the mid-points of the intercepts of the tangents totheellipse x2/a2 + y2/b2 = 1 = 1, intercepted between the axes, is a2/x2 +b2/y2 = 4.

Solution:

The tangent to the ellipseat any point (a cos, b sin)(x cos)/a+(y sin )/b = 1.

Let it meet the axes in P and Q, so that P is (a sec, 0)

and Q is (0, b cosec). If (h, k) is the mid-point of PQ, then h = (a sec)/2.

=> cos = a/2h and k = (b cosec)/2 => sin = b/2k.

Squaring and adding, we get a2/4h2 +b2/(4k2 ) = 1

Hence the locus of (h, k) is a2/x2 +b2/y2 = 4.

Illustration:

Prove that the product of the lengths of the perpendiculars drawn from the foci toany tangent to the ellipse x2/16+y2/9 = 1 is equal to 9.

Solution:

For the given ellipse a = 4, b = 3 and hence 9 = 16 (1 - e2)

=> e = 7/4. The foci are thus located at (7,0) and (-7,0).

Equation of a tangent to the given ellipse is

y = mx + (16m2+9) (as a = 4, b = 3).

Lengths p1 and p2 of the perpendiculars drawn from the foci are

p1= ((16m2+9)+7 m)/(1+m2 ) and p2= ((16m

2+9)-7 m)/(1+m2)

=> p1p2 = (16m2+9-7m2)/(1+m2) = 9(1+m2)/(1+m2) = 9.

Tangent and NormalNote:

The product of lengths of the perpendiculars drawn from the foci to any tangent tothe ellipse x2/a2 +y2/b2 = 1 is b2.

Equation of the Normal to an Ellipse


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The normal to a curve is a line perpendicular to the tangent to curve through eh pointof contact.

.. The slope ofnormal at point (x1, y1) = a2y1/b

2x1 and so its equation is (x-x1)/((x1/a

2))=(y-y1)/((y1/b2 ) ).

* To find the equation of the normal to the ellipse x2/a2 +y2/b2 = 1 at (x1, y1):

Equation of the tangent at (x1, y1) is (xx1)/a2 +(yy1)/b

2 = 1

=> Slope of the normal is a2/b2 +y1/x1; => equation of the normal is

y - y1 = a2/b2 +y1/x1 (x - x1) =>(x-x1)/(x1/a

2)=(y-y1)/(y1/b2 ).

* Equation of the normalat (a cos, b sin) is (x-a cos)/((a cos)/a2) = (y-b sin

)/((b sin)/b2).

=> ax sec - by cosec = a2 - b2.

ax1sec - by1cosec = a2 - b2.

or ax1((1+t2)/(1-t2)) - by1 ((1+t

2)/2t) = a2 - b2, where t = tan /2.

On simplification, this equation gives

by1t4 + 2(ax1 + a

2 -b2)t3 + 2(ax1 - a2 + b2)t - by1 = 0.

This is a 4th degree equation in t which gives, in general, four values fo t. Hence from afixed point four normals can be drawn to the given ellipse.

Illustration:

If the normals to the ellipse x2/a2 + y2/b2 = 1 at the points (x1, y1), (x2, y2) and (x3,

y3) are concurrent, prove that = 0.

Solution:

The equation of the normal to the given ellipse at (x1, y1) is

a2xy1 - b2yx1 - (a

2 - b2)x1y1 = 0. ...... (1)

Similarly the normals at (x2, y2) and (x3, y3) are

a2xy2 - b2yx2 - (a

2 - b2)x2y2 = 0. ...... (2)

a2xy3 - b2yx3 - (a

2 - b2)x3y3 = 0. ...... (3)


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Eliminating a2x, b2y and (a2 - b2) from (1), (2) and (3), we find that the three lines are

concurrent if = 0.

Tangent and NormalIllustration:

If the normals at the end of a latus rectum of the ellipse x2/a2 +y2/b2 = 1 passesthrough the extremity of a minor axis, prove that e4+e2-1=0.

Solution:

Equation the normal to te given ellipse at

(ae,b2/a) is (x-ae)/(ae/a2) = (y-b2/a)/(b2/ab2).

Since it passes through (0, - b).

- a2 = - ab - b3

=> a2 = ab + a2 (1 - e2)

or b = ae2 => b2 = a2e4

or a2 (1 - e2) = a2e2 a4 + e2 - 1 = 0.

Enquiry: How many tangents can be drawn from a point on an ellipse?

We know that y = mx + (a2 m2+b2) is a tangent to the ellipse x2/a2 +y2/b2 = 1 forreal values of m. If this tangent passes through a point (x1, y1) we have y1 = mx1 +((a2 m2+b2)).

or m2(x12 - a2) - 2x1y1m + (y1

2 - b2) = 0, which being a quadratic equation in m givestwo values of m. Thus from a point two tangents corresponding two values of m) can

be drawn to an ellipse.

Pair of Tangents and Chord of Contact

From a fixed point (x1, y1) in general two tangents can be drawn to an ellipse. Theequation of the pair oftangents drawn to the ellipse x2/a2 +y2/b2 = 1 is given by(x2/a2 +y2/b2-1)((x1

2)/a2 +(y12)/b2-1) = (xx1/a

2 + yy1/b2-1)2.


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Tangent and Normal

In symbols we write SS1 = T2

, where

S x2/a2 +y2/b2 = 1, S1 (x12)/a2 +(y1

2)/b2 1 and T xx1/a2 + yy1/b

2 1

* If from the point P(x1, y1) tangents PQ and PR be drawn to the ellipse x2/a2+y2/b2 =

1, then the line joining the points of contact Q and R is called the chord of contact.Equation of the chord of contact is xx1/a

2 + yy1/b2 - 1 = 0 or T = 0.

* Equation of the Chord Joining the Points ( cos, sin),( cos, sin) is

x/a cos((+)/2)+ y/b sin((+)/2) = cos((-)/2)

* Equation of a chord which is bisected at the point (x1, y1) is xx1/a2 + yy1/b

2 -1 =(x1

2)/a2 +(y12)/b2 - 1 or = S1

* Length of the chord ... (from package).

To find the length of the chord intercepted by the ellipse x2/a2 +y2/b2 = 1 onthe straight line y = mx + c.

Points of intersection of the ellipse and the line are given by x2/a2 +(mx+c)2/b2 = 1

i.e. (a

2

m

2

+ b

2

)x

2

+ 2a

2

cmx + a

2

(c

2

- b

2

) = 0 ...... (1)

Therefore the straight line meets the ellipse in two points (real, coincident orimaginary).

If (x1, y1) and (x2, y2) be the points of intersection, the length of the chord is

((x1-x2 )2+(y1-y2)

2)=(1+m2 )= |x1 - x2| ...... (2)

(since y1 - y2 = m (x1 - x2))

where x1 and x2 are the roots of the equation (1), and

x1 + x2 = -(2a2cm)/(a2 m2 + b2), x1 x2 (a

2 (c2 - b2))/(a2 m2 + b2) so that


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(x1 - x2)2 = (x1 + x2)

2 - 4x1 x2 = (4a2 c2 m2)/(a2 m2 + b2)2 -(4a2(c2 - b2))/((a2m2

+ b2))=(4a2 b2 (a2 m2 + b2 - c2))/(a2 m2 + b2)2.

Hence the length of the chord is (((1+m2)4a2 b2 (a2 m2 + b2 - c2))/(a2 m2 + b2)2).

i.e. 2ab/(a2 m2 + b2)2) ((1+m2))(a2) m2)+b2)-c2))).

=> e cos(+)/2 = cos(-)/2

=> e [cos /2 cos /2 + sin /2 sin /2] = cos /2 cos /2 + sin /2 sin /2

=> e [1 - tan /2 tan /2] = 1 + tan /2 tan /2 => tan /2 . tan/2 = (e -1)/(e+1).

If the chord passes through (-ae, 0) then tan/2 . tan/2 = (e+1)/(e-1).

Tangent and NormalIllustration:

The tangent and normal at a point P on an ellipse meet the minor axis at A and B.Prove that AB subtends a right angle at each of foci.

Solution:

The equations of the tangents and normal at a point P(x1, y1) on the ellipse x2/a2+

y2/b2 = 1 are

xx1/a2 + yy1/b

2 = 1 ...... (1)

and (x-x1)/((x1/a2))+(y-y2)/((y1/b

2)) = 1 ...... (2)

Also the minor axis is y-axis i.e. x = 0

Solving (1) and x = 0, we have A (0, b2/y1)

Solving (2) and x = 0, we have B (0, y1 - a2y1/b

2)

Let S(ae, 0) be one of the foci of the ellipse. Then the slope of SA = ((b

2

/y1) - 0)/(0-ae)= b2/aey1 = m1 (say)

And the slope of SB = ([y1 (a2/b2 ) y1 ]-0)/(0-ae)

= y1/ae ((a2-b2 ))/b2= (y1 a

2 e2)/(aeb2) [. b2 = a2(1 - e2)]

= aey1/b2 = m2 (say)

Evidently m1m2 = -1 => SA | SB

i.e. AB subtends a right angle at S(ae, 0). Similarly we can show that AB subtends a

right angle at the other focus S'(-ae, 0)


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Illustration:

Prove that the locus of the middle points of the portion oftangent tothe ellipsex2/a2 +y2/b2 = (a+b) between the axis is the curve a/x2 + b/y2 = 4/((a+b)).

Tangent and NormalSolution:

Any tangent to the ellipse x2/a2 +y2/b2 = 1 , is

y = mx + (a2 m2 + b2)

Therefore the tangent to given ellipse is

y = x + (a(a+b) m2+b(a+b)) ...... (1)

(1) Meets x-axis at {-(a(a+b) m2+ b(a+b) )/m,0}

(2) Meets y-axis at {0,(a(a+b) m2+b(a+b))}

Let (x1, y1) be the mid points of the portion of the tangent intercepted between theaxes, then

x1 = 1/2 [{-(a(a+b) m2 + b(a+b))/m} + 0]

y1= 1/2 [0+{(a(a+b) m2+b(a+b))} + 0]

=> 4x12 = (a(a+b) m2+ b(a+b))/m2

and 4y12 = a(a + )m2 + b(a + b)

Eliminating m, we get

a(a+b)/(4x12) - b(a+b)/(4y1

2) = am2/(am2+b) + b/(am2+b) = 1

.. The locus of (x1, y1) is a/x2 + b/y2 = 4/(a+b) = 1

Illustration:

If PS1Q and PS2R be two focal chord of the ellipse whose two foci are S1 and S2and theeccentric angle of P is '' then show that the equation of chord QR is x/a cos +y/b.(1+e2)/(1-e2) sin + 1 = 0.

Solution:

Let Q be (a cos , b sin ) and R be (a cos , b sin ) then the equation of the chord QRis

x/ cos(+)/2 + y/b sin(+)/2 = cos(-)/2

which on simplifying becomes


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x/a (1-tan /2 tan /2)+y/b (tan /2+tan /2) = 1 + tan /2 tan /2 ..... (1)

Also, PQ and PR are focal chord thus

tan /2 tan /2 = (e-1)/(e+1) and tan /2 tan /2 = (e+1)/(e-1)

(From previous illustration)

On substituting the values of tan /2 and /2 in (1), we get

Tangent and Normal

Illustration:

Show the locus of middle points chord of the ellipse x2/a2 +y2/b2 = 1 which subtendright angle at the centre is x2/a4 + y2/b4 =(1/a2 +1/b2 ) (x2/a2 + y2/b2)2.

Solution:

Let (x1, y1) be the middle point of chord PQ, then its equation is

T = S1 or 1/a

2

+ yy1/b

2

=(x1/b

2

)/a

2

+(y1/b

2

)/b

2

...... (1)

Since the origin 'O' is the centre so the equation of pair of lines OP and OQ can beobtained by homogenizing the equation of the ellipse x2/a2 +y2/b2 = 1, with the help of(1), thus

x2/a2 +y2/b2 = (1)2 =

or ((x12)/a2 +(y1

2)/b2)2 (x2/a2 +y2/b2 ) = (x2 x12)/a4 +(y2 y1

2)/b4 + (2xyx1 y1)/(a2b2).

It represents a pair of perpendicular lines if


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1/a2 ((x12)/a2 +(y1

2)/b2 )2 - (x12)/a4 + 1/b2 ((x1

2)/a2 +(y12)/b2)2-(y1

2)/b4 = 0.

or, (x12)/a4 +(y1

2)/b4 = (1/a2 +1/b2)(x2/a4 +y2/b4)2.

So the locus of (x1, y1) is

x2/a4 + y2/b4 = (1/a2 + 1/b2 ) (x2/a2 +y2/b2 )2.

Propositions on EllipseAuxiliary circle of an ellipse

Auxiliary circle of an ellipse which is a circle described on the major axis ofanellipse as its diameter.

Let the ellipse be

x2/a2 + y2/b2 =1 ...... (1)

Then the equation of its auxiliary circle is x2 + y2 = a2 ...... (2)

Take a point P(x1, y1) on (1).

Through P, draw a line perpendicular to major axis intersecting major axis in N andauxiliary circle in P'.

Let P' be (x1, y2). Then we have (x12)/a2 + (1

2)/b2 = 1 and 12 + y12 = a2

From these two relations, we get:

(y22)/(y1

2)=(a2-12)/(b2 {1-(x1

2/a2)}) = a2/b2

=> y2/1 = a/b

Now, let OP' make an angle f with the major axis of the ellipse (1), P' being thecorresponding point of P on the auxiliary circle of the ellipse. Then is called theeccentric angle of the point P. From the figure it is evident that if (x1, y1) are the co-ordinates of P, then x1= ON = OP' cos = a cos. Also P(x1, y1) is a point on (1).

So solving(x12)/a2 +(12)/b2 =1 for y1, we get y1= b sin


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Therefore (a cos , b sin ) are the parametric co-ordinates of any point P ontheellipse, where is the eccentric angle of P.

Propositions on Ellipse:Note:

1. Tangent to the ellipse at the point:

The equation of the line PQ reduces to that of tangent when x/a cos + y/b sin =1 ...... (a)

2. Normal to the ellipse at the point 'f':

Equation of normal can be derived by using the formula of equation of straight linepassing through point (a cos, b sin) and perpendicular to tangent (a) i.e. (y - b sin )= (a sin /b cos ) (x - a cos ).

=> a x sec - b y cosec = a2 - b2

Diameter of an ellipse

The locus of the middle points of a system of parallel chords of an ellipse is calledthe diameter of theellipse.

Let y = mx + c ...... (1)

be the equation of a system of parallel chords of the ellipse x2/a2 + y2/b2 =1 ...... (2)


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In (1) m is constant and c varies from chord to chord. Let K (x1, y1) be the midpoint of achord PQ of this system.

Eliminating y between (1) and (2) we get

(a2m2 + b2)x2 + 2a2mcx + a2(c2 - b2) = 0 ...... (3)

=> x2 + x3 = (-2a2 mc)/(a2 m2+b2 )

But x1 = (x2+x3)/2 =(-a2 mc)/(a2 m2+b2)

Or c = (-x(a2 m2+b2))/(a2 m2)

Also K (x1, y1) is a point on (1) so

y1 = mx1 + c => y1= -b2x1/a

2m

.. The locus of K (x1, y1) is y = -b

2

x/a

2

m which is a diameter of the ellipse x

2

/a

2

+y2/b2 = 1.

Propositions on Ellipse:

Note:

1. Conjugate Diameters: Two diameters of anellipse which bisects chords parallel to each other are called

conjugated diameters. Therefore the diameters y = mx and y = m1x of the ellipse x2/a

2+y

2/b

2= 1 are conjugate

if mm1 = -b2/a

2.

2. In an ellipse, the major axes bisects all chords parallel to the minor axes and vice-versa, therefore major axesand minor axes of an ellipse are conjugate diameters but they do not satisfy the condition

mm1 = -b2/a

2and are the only perpendicular conjugate diameters.

3. Equi-conjugate diameter:

If the length of two conjugate diameters ofellipse be equal then they are called equi conjugate diameters.

The equation of equi conjugated diameters are x2/a

2 y

2/b

2= 1.


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4. The eccentric angles of the ends of a pair of conjugate diameters of anellipsediffer by a right angle i.e., if

one end of a diameter (PQ) is

P(a sin , b cos ).

5. The sum of the squares of any two conjugate semi-diameters of anellipse is constant and equal to the sum

of the squares of the

semi-axis of the ellipse i.e. OP2

+ OP2

= a2

+ b2

.

6. The produce of the focal distances of a point on an ellipse is equal to the square of the semi-diameters, which

is conjugate to the diameter through the point.

7. The tangents at the ends of a pair of conjugate diameters of an ellipse form a parallelogram and the area of

the parallelogram is constant and is equal to the product of the axis i.e. equal to 4ab.

Propositions on Ellipse:Director circle of an ellipse

The director circle is the locus of the point of intersection of pair of perpendiculartangents to an ellipse.

Two perpendicular tangents ofellipse x2/a2 + y2/b2 = 1 are

y - mx = (a2 m2+b2) ...... (1)

and my + x = (a2+b2 m2) ...... (2)

To obtain the locus of the point of intersection (1) and (2) we have to eliminate msquaring and adding (1) and (2), we get

(y - mx)2 + (my + x)2 = (a2m2 + b2) + (a2 + b2m)

=> x2 + y2 = a2 + b2, which is the equation of the director circle.

Enquiry: How to find equation of a chord of an ellipse whose mid point is (x1,y1)?

Let the ellipse be x2/a2 + y2/b2 = 1 ...... (1)

Any line through (x1, y1) is y - y1 = m(x - x1) ...... (2)

Eliminating y between (1) and (2) we get


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x2 (b2 + a2m2) + 2ma2 (y1 - mx1) x + a2 [(y1 - mx1)

2 - b2] = 0

If x2 and x3 are the roots of this equation then

x2 + x3 = -2ma2 (y1 - mx1)/(b

2 + a2m2)

But (x1, y1) the mid point of the chord.

.. x1 = (x2+x3)/2=(-ma2 (y1-mx1 ))/(b

2+a2 m2)

=> m = (-b2 x1)/(a2 y1)

.. From (2) and (3) the equation of the chord where mid point is (x1, y1) is

y - y1 = ((-b2 x1)/(a

2 y1)) (x - x1)

=> xx1/a2 + yy1/b

2 =(x12)/a2 +(y1

2)/b2

Or T = S1 where T = xx1/a2 + yy1/b

2 -1

and S1 = (x12)/a2 + (y1

2)/b2 -1

Propositions on Ellipse:Pole and polar of an ellipse

The locus of the points of intersection of tangents drawn at the point extremities of thechords passing through a fixed point is called the polar of that fixed point and the fixedpoint is called the pole.

Equation of the polar of P(x1, y1) with respect to the ellipse

x2/a2 + y2/b2 = 1 is

xx1/a2 + yy1/b

2 =1

Illustration:


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How to find out pole of the line lx + my + n = 0 w.r.t. the ellipse x2/a2+y2/b2 = 1.

Solution:

Let (x1, y1) be the pole of line lx + my + n = 0. ...... (1)

w.r.t. the ellipse x2/a2 + y2/b2 = 1 ...... (2)

Now the polar of (x1, y1) w.r.t. (2) is

xx1/a2 + yy1/b

2 = 1 ...... (3)

Since (1) and (3) represents the same polar, so comparing them we have

(x1/a2)/l+(y1/b

2)/b2 = (-1)/n

or x1 = (-a2 l)/n y1 = (-b

2 m)/n

.. The required pole is ((-a2 l)/n,(-b2 m)/n)

Solved Examples of Ellipse:Example 1:

Find the points on the ellipse x2 + 3y2 = 6 where the tangent are equally inclined tothe axes. Prove also that the length of the perpendicular from the centre on either ofthese tangents is 2.

Solution:

The given ellipse is x2 + 3y2 = 6

Or x2/6+y2/2 = 1 ...... (1)

If the coordinates of the required point on the ellipse(1) be (6 cos ,2 sin ) thenthe tangentat the point is x/6 cos + y/2 sin = 1 ...... (2)

Slope of (2) = (-cos )/62/(sin )=(-2)/6 cot

As the tangent are equally inclined to the axes so we have

-1/3 cot = + tan 45o = + 1

.. tan = + 1/3

The coordinates of the required points are

(63/2, 21/2) and (63/2, +-21/2)

= ((32)/2,1/2) and ((32)/2,+-1/2)

Again the length of perpendicular from (0, 0) and (2),


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= (6.2)/(2 cos2 + 6 sin2)=(23)/((2,3/4)+(6.1/4) )=(23)/3 = 2

Example 2:

If P be a point on the ellipse x2/a2 + y2/b2= 2/c whose ordinate is 2/c, prove that theangle between the tangent at P and SP is tan-1 (b2/ac), where S is the focus.

Solution:

The given ellipse x2/a2 + y2/b2 = 2/c ...... (1)

If (x',(2/c)) be the coordinates of the given point P on theellipse (1).

Then the tangent at P will be:(xx')/a2 +(yy')/b2 = 1

(xx')/(a2(2/c) )+(yy' ((2/c)))/(b2 (2/c))= 1

The slope oftangent at P is (-b2 (2/c) x')/(a (2/c)(2(2/c)))= m1 (say)

If S be the focus, then slope of PS = = y'/(x'+ae)=(2/c)/(x'+a(2/c)e)= m2 (say)

If angle between the focal distance SP and tangentat P is , then

tan = (m2-m1)/(1+m2 m1).

(a2 (2/c)+b2 x'2+ aeb2x' (2/c))/((a2 x'+a3e(2/c)-b2x')(2/c))

point (x',(2/c)) lies onellipse (1), we have

b2 x'2 + a2(2/c)=(2a2 b2)/c

and . a2 - b2 = a2 e2 so we have

(-(2a2 b2)/c + aeb2x' (2/c))/((a2 e2 x'+a2e(2/c)) (2/c)) = ((2/c) ab2((2/c)

a+ex'))/(a2e((2/c a+ex')) (2/c)) = b2/ae.

.. = tan-1 b2/ae. Hence proved.

Example 3:

If P, Q, R are three points on the ellipsex2/a2 +y2/b2= 1 whose eccentric angles are , and then find the area of PQR.

Solution:

The coordinates of the gives points P, Q, R on the ellipse x2/a2 +y2/b2 = 1, will be (a cos, b sin ), (a cos , b sin ) and (a cos , b sin ). Area of triangle PQR formed bythese points

= 1/2 [x1y2 - x2y1 + x2y3 - x3y2 + x3y1 - x1y3]


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= 1/2 [ab cos sin - ab sin cos + ab cos sin - ab sin cos + ab in cos - ab cos sin ]

= 1/2 ab [2 sin(-)/2 cos (-)/2 + 2 sin (-)/2 cos (-)/2 + 2 sin (-)/2 cos(-)/2]

= ab sin(-)/2 (cos(++2)/2 cos(-)/2)

= 2ab sin (-)/2 cos (-)/2 cos (-)/2.

Example 4:

Find the locus of the extremities of the latus recta of all ellipses having a given majoraxis 6a.

Solution:

Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be(x, y) then x = CS = 3 ae ...... (1)

And y = SL = b2/3a =(9a2(1-e2))/3a = 3a (1 - e2) ...... (2)

Eliminating the variable 'e' from (1) and (2) we get eh locus of L.

Hence putting the value of e from (1) and (2), we get

y = 3a(1-x2/9a2)

=> x

2

= 9a

2

- 3ay

=> x2 = 3a(3a - y), which is clearly a parabola. Similarly we can show that the locus ofL' is x2 = 3ay(y + 3a) which is again a parabola.

Example 5:

Find the equation of the normals at the end of the latus rectum ofthe ellipsex2/a2 +y2/b2 = c2 and find the condition when each normal through one endof the minor axis.

Solution:

The ellipse x2/a2 + y2/b2 = c2

=> x2/(a2 c2+y2/(b2 c2) = 1 ...... (1)

Then the end point of the latus rectum is (ace,(b2 c)/a)

The normal at this point will be

(x-ace)/(ace/(a2 c2)) = (y-b2 c/a)/((b2 c/a)/(b2 c2))

=> ((x-ace))/e ac = (y-(b

2

c)/a) ac


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=> (x-ace)/e = y-(b2 c)/a

If this normal passes through (0, - bc), then, we have

(-ace)/e = -bc-b2/a c

=> a = b + a(1 - e2)

=> b - ae2 = 0

=> b/1-e2 => b2/a2 = e4

=> e4 + e2 = 1. This is required condition.

Example 6:

The circle x2 + y2 = 4 is concentric with the ellipse x2/7 + y2/3 = 1; prove that thecommon tangent is inclined to the major axis at an angle 30o and find its length.

Solution:

The ellipse x2/7+y2/3=1 ...... (1)

The equation to the circle is

x2 + y2 = 4 ...... (2)

As the line y = mx + (a2 m2+b2)

i.e. y = mx + (7m2+3) ...... (3)

is always the tangent on the ellipse. If this is also a tangent on the circle (2) thenlength of perpendicular from the centre (0, 0) on the line (1) must be equal to radius ofcircle i.e. 2.

Hence, (7m2+3)/(1+m2) = 2

=> 7m2 + 3 = 22 (1 + m2)

=> 7m2 - 4m2 = 4 - 3

=> m2 = 1/3

=> m = + 1/3

Hence the common tangent to the two curves is inclined at an angle of tan -1(+1/3)i.e. 30o to the axis.

Note:


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We can also prove the above result by using the fact that the line = mx + (7m2+3) willbe tangent to x2 + y2 = 4 if discriminent of x2+ (m + (7m2+3))2 = 4 is zero.

Let P and Q be the points of contact of the common tangent with ellipse and circlerespectively and O be the common centre of the two, then PQ = (OP2-OQ2)[CPQ =90o]

The coordinates of P are [(-a2m)/(a2 m2 + b2), b2/(a2 m2 + b2)]

i.e. [(-7/3)/(16/3),3/(16/3)]

i.e.[(-7)/4,(33)/4]

and coordinate of O are (0, 0)

So, OP = (((-7)/4)2+ ((33)/4)2) = (19/4)

As OQ = r = 2

.. PQ = (OP2-OQ2) = (19/4-4) = 3/2

Example 7:

If q be the angle between CP and normal at point P, on the ellipse a2 x2 + b2 y2 = 1,then find out tan and prove that its greatest value is (b2-a2)/2ab. C is centreofellipse and P is any point on ellipse.

Solution:

The equation of the ellipse be

x2/(1/a2)+y2/(1/b2) = 1 ..... (1)

If be the angle between the normal at P = (1/a cos ,1/b sin and PC where C is thecentre of the ellipse given by (1) equation to the normal PG is

x/a sec - y/b cosec = 1/a2 -1/b2

=> bx sec - ay cosec = (b2 - a2)/ab

Its slope = (b sec )/(a cosec ) = b/a tan = m1 (say)

The slope of PC = (1/b sin )/(1/a cos ) = a/b tan = m2 (say)

tan = (m1-m2)/(1+m1 m2) = (a/b tan -a/b tan )/(1+(b/a) tan (a/b) tan ) =((b2-a2)tan )/ab(1-tan2)

(b2-a2)/2ab.(2tan)/(1-tan2)

tan = (b2 - a2)/2ab sin 2

The value of tan will be maximum when sin 2 is maximum, sin 2 is maximum i.e.sin 2 is 1. Therefore the greatest value of tan is (b2-a2)/2ab.


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Example 9:

Find the locus of the point of intersection of the two straight lines (x tan )/a - y/b + tanx = 0 and x/a+(y tan )/b where a is fixed angle. Also find the eccentric angle of thepoint of intersection.

Solution:

Equation of the lines are given as

(x tan )/a-y/b + tan = 0 ...... (1)

x/a+(y tan )/b - 1 = 0 ...... (2)

To find the locus of the point of intersection, we have to eliminate the variable 'tan a'from (1) and (2), so by (2),

y/b=1/(tan ) (1-x/a) and by (1)

y/b = tan (1+x/a)

Multiplying we get

y/b)2-(1-x/a)(1+x/a)

=> x2/a2 +y2/b2 = 1

This is the equation of an ellipse

Again solving (1) and (2), we get

x = a(1-tan2)/((1+tan2))

x = a(1-tan2)/(sec2)

Let the abscissa of the point of intersection be a cos f, then

x = a cos = a(1-tan2)/(sec2)

=> cos = (1-tan2)/(sec2)

=> (1-cos )/(1+cos )=(sec2 -(1-tan2))/(sec2 +(1-tan2 ) )

(By components & dividendo)

= (2 tan2)/2 = tan2

=> (2 sin2/2)/(2 cos2) = tan2

=> tan2/2 = tan2

=> tan /2 = tan


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Hence = 2

Example 10:

If TP and TQ are perpendiculars upon the axes from any point T on the ellipsex2/a2+y2/b2 = 4. Prove that PQ is always normal to fixed concentric ellipse.

Solution:

The ellipse is given by x2/(4a2 )+ y2/(4b2 ) = 1 ...... (1)

If the co-ordinates of T on the ellipse be (2a cos f, 2b sin f) an TP and TQperpendiculars on x-axis and y-axis respectively, the co-ordinates of P and Q will be (2acos f, 0) and (0, 2b sin f) respectively.

Now equation to PQ is

y - 0 = (2b sin -0)/(0-2a cos )(x - 2a cos )

=> x/(2a cos )+y/(2b sin ) = 1

=> x/a sec + y/b cosec = 2 ...... (2)

Now equation to the normal at point (A cos , B sin ) with respect to any otherconcentric ellipse

x2/A2 +y2/B2 = 1 is

Ax sec - By cosec = A

2

- B

2

...... (3)

As (2) and (3) are similar, on comparing then we have,

A/(12a)+B/(12b) = A2 - B2 ...... (4)

Solving the two equations given by (4), we get

B = (2a2b)/(a2-b2) and A = (-2 ab2)/(a2-b2 )

So the line (2) i.e. x/2a sec + y/2 cosec = 1 is a normal to thefixed ellipsex2/A2 +y2/B2.

Where A = (-ab2)/(a2-b2)and B =(a2 b)/(a2-b2).

Example 11:

If the straight line y = 2x + 2 meet the ellipse 2x 2 + 3y2 - 6, prove that equation to thecircle, described on the line joining the points of intersection as diameters, is 7x 2+ 7y2 +12x - 4y - 5 = 0.

Solution:

The line is given as


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y = 2x + 2 ...... (1)

and the ellipse is given as

x2/a2 +y2/b2 = 1 ...... (2)

Solving (1) and (2), we gets

x2/3+(2x+2)2/2 = 1

=> x2/3 + 2(x2 + 2x + 1) = 1

=> 7x2 + 12x + 6 = 3

=> 7x2 + 12x + 6 = 3

Let x1 and x2 be the two roots of this equation.

x1 + x2 = -12/7 ...... (3)

and x1x2 = 3/7 ...... (4)

Let y1 and y2 be the corresponding ordinates for the abscissa x1 and x2; so the co-ordinates of the points of intersection will be (x1, y1) and (x2, y2). As these lie on line

y = 2(x + 1)

We have y1 = 2(x1 + 1) and y2 = 2(x2 + 1)

Where y1 + y2 = 2(x1 + x2) + 4 ...... (5)

y1y2 = 4(x1 + 1) (x2 + 1) ...... (6)

The equation to the circle drawn with the line joining (x1, y1) and(x2, y2) as diameter is

(x - x1) (x - x2) + (y - y1) (y - y2) = 0

=> x2 + y2 - x(x1 + x2) - y(y1 + y2) + x1x2 + y1y2 = 0

=> x2 + y2 - x(x1 + x2) y[2(x1 + x2) + 4] + x1x2 +......+ 4[x1x2 + (x1 + x2) + 1] = 0

(By (5) and (6))

Putting the values from (3) and (4)

x2 + y2 - x (-12/7) - y[2(-12/7)+4] + 3/7 + 4[3/7+(-12/7)+1] = 0

= 7x2 + 7y2 + 12x - 4y - 5 = 0. Hence proved.

Example 12:


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If the product of the perpendiculars from the foci upon the polar of P be constant andequal to c2. Find the locus of P.

Solution:

Suppose the equation to the ellipse x2/a2 + y2/b2 = 1. The co-ordinate of foci are (ae, 0)and (-ae, 0).

Let the co-ordinates of P be (h, k). Then polar of P is

xh/a2 +yk/b2 = 1

Or b2xh + a2yk - a2b2 = 0 ...... (1)

If P1 and P2 be the lengths of the perpendiculars on the line (1) from (ae, 0) respectivelyare

P1 = (b2 hae-a2 b2)/(b2 h2 + a4 k2)

And P2 = (-a2 b2-b2hea)/(b2 h2+a4 x2)

.. P1P2 = (-b2 h2 a2 e2+a4 b4)/(b4 h2+a4) = c2 (By hypothesis)

=> a4b4 - b4h4a2e2 = c2b4h2 - c2a4k2

=> b4h2 (c2 + a2e2) + c2a4k2 = a4b4

Generalizing the locus of the point P(h, x) is

b4 x2 (c2 + a2 e2) + c2 a4 y2 = a4 b4

Example 13:

Chords of ellipse x2/a2 +y2/b2 = 1 always touch another concentric ellipse x2/2+y2/2= 1, show that the locus of their poles is (2 x2)/a2+ (2 y2)/b2 = 1.

Solution:

Let (x1, y1) be the pole of a chord of the ellipse

x2/a2 +y2/b2 = 1 ...... (1)

Then the equation of this cord is the same as the polar of (x1, y1) with respect to (1)

i.e. xx1/a2 + yy1/b

2 = 1 ...... (2)

If (2) touches the ellipse x2/2 + y2/2 = 1, then

(b2/y1)2= 2{-{b2x1/a

2y1}}2+ 2

=> b4/y21= (2b4x21/a

4y21) + 2

=> (2 x12)/a4+(2 y1

2)/b4 = 1


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.. The locus of (x1, y1) is (2 x2)/a2+(2 y2)/b4 = 1

Hence proved.

Example 14:

If the straight line y = x tan + ((a 2 tan2 +b2)/2), being the angle of inclination,intersects the ellipse x2/a2 +y2/b2 = 1. Then prove that the straight lines joining thecentre to their point of intersection are conjugate diameters.

Solution:

The equation of the ellipse be

x2/a2 +y2/b2 = 1 ...... (1)

and equation to the line is given as

y = x tan + ((a2 tan2 +b2)/2), being angle of inclination.

We can write this equation as,

y = mx + ((a2 m2 +b2)/2)

=> ((y-mx)/2)/(a2 m2 + b2 ) = 1 ...... (2)

To get the equation to the lines joining the point of intersection to the origin, making (1)homogeneous with the help of (2), we have

x2/a2 + y2/b2 = [((y-mx)/2)/(a2 m2 + b2 )]2

= 2((y2 + m2 x2 - 2mxy))/(a2 m2 + b2)

=> (b2 x2 + a2 y2) (a2m2 + b2) = 2a2 b2(y2 + m2x2 - 2mxy)

y2 a2(a2m2 - b2) + 4m2 - b2xy - b2x2 (a2m2 - b2) = 0

=> y2 + (4 mb2)/((a2 m2+b2)) xy -b2/a2 x2 = 0 ...... (3)

This equation represents two straight lines y = m1x and y = m2x then the combinedequation will be y2 - (m1 + m2)xy + m1m2x

2 = 0.

Comparing (3) and (4); we get

m1m2 = -b2/a2

which is the condition of diameter to be conjugate. Hence the lines are the conjugatediameters.

Example 15:


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The eccentric angles of two points P and Q on the ellipse 1, 2. Find the area of theparallelogram formed by the tangents at the ends of diameters through P and Q.

Solution:

The ellipse is

x2/a2 +y2/b2 = 1

Equation to the tangent at the points P and Q are

x/a cos 1+ y/b sin 1 = 1 ...... (1)

and

x/a cos 2+ y/b sin 2 = 1 ...... (2)

Solving (1) and (2), we will have the coordinates of the point of intersection. Multiplying(1) by sin 2and (2) by sin 1 and subtracting, we get

x/a (sin 2cos 1 - cos 2sin 1) = sin 2 - sin 1

=> x/a sin (2 - 1) = 2 sin (2 - 1)/2cos (2+ 1)/2

.. x = a (cos((1 - 2)/2))/(cos(1 - 2)/2); y = b (sin(1+ 2)/2)/(cos (1 - 2)/2)

Above are co-ordinates of the point of intersection L of tangents at p and Q, i.e. at1and 2. Putting 1= p + 1 in above we get the co-ordinates of the point of

intersection M of tangent at Q and P' as

Area of the parallelogram LMNO = 4CLM

= 4 . 1/2 (x1 y1 - x2 y2)

= 2 ab/(sin (1 - 2)/2 cos (1 - 2)/2) [ - cos2(1+ 2)/2-sin (1+ 2)/2]

= (-4ab)/(sin(1 - 2))

= - 4ab cosec (1 - 2) Area can't be (-) ve

So the area = 4ab cosec (1 - 2)

Basic Concepts of Ellipse

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