Ellipse
01. Introduction 02. Tangents
03. More on Tangents and Chords 04. Normals
CONCEPT NO
CONCEPT NO
CONCEPT NO
Ellipse
Parabolas, which we studied in the last chapter, are conic sections with eccentricity e = 1. In this section, we study ellipses, which have e < 1. This means, by definition, that 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. In fact we will soon discover 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
Fig - 1 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
e
Q L =QN =
Draw a line through F perpendicular to the directrix L, which meets the directrix at M. On this line, we can always find two points, say P and P', which divide the line segment MF internally and externally respectively in the ratio 1 : e. Fig - 2 y x M Directrix L P F P'
We can always find two points and ' such that
P P
PF PM=
P F' P M' = e
Thus, by definition, P and P ' will both lie on the ellipse.
We will now introduce co-ordinates into this geometry. We assume the x-axis to coincide with the horizontal line and 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): Fig - 3 y x P -a, ( 0) F P -a, '( 0) M L Since PF e PM = and , P F e P M ′ = ′ we have ( ) PF+P F′ =e PM +P M′ ( ) PP′ e PM P M′ ⇒ = + 2a e(2PM PP′) ⇒ = + =2ePM +2ae (1 ) a PM e e ⇒ = −
This gives PF=ePM =a(1−e) and OM OP PM a. e
= + = Thus the co-ordinates of F are (–ae, 0) and that of
M are a, 0 . e
−
Till now we have assumed F to be the focus of the ellipse and L to be the directrix. However, by the symmetry inherent in Fig -3, you might be able to infer that there should exist another focus F ' and directrix L ', to the right of O as shown below: Fig - 4 y x M P F O (– ,0)a (–ae,0) (aeF ',0) P ' ( ,0)a M ' a e,0) ( -a e,0) ( L L'
To find the equation of the ellipse we assume the co-ordinates of any point Q lying on the ellipse as (x, y): Fig - 5 y x F(–ae,0) F ' (ae,0) -a e,0) ( L L' N Q x, y( ) N ' a e,0) ( We thus have QF e QN = 2 2 2 2 (x ae) y e a x e + + ⇒ = + 2 2 2 2 2 2 2 2 2 x 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 y a a e ⇒ + = − 2 2 2 2 1 x y a 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 ' and another directrix L '. In fact, you are now urged to verify the truth of this explicitly by proving that for any point Q(x, y) lying on the ellipse as in Fig - 5, we always have
QF QF e QN QN ′ = = ′
We finally plot the ellipse taking help of the equation obtained. Fig - 6 y x P ' L L' N Q N ' M ' F ' F O P 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, R≡(0, )b and R′ ≡(0,−b).
PP', whose length equals 2a, is termed the major-axis of the ellipse, while RR ', whose length equals 2b, is termed the minor-axis 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, MM 2a.
e ′ =
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 e QN QN ′ = = ′ ( ) QF QF′ e QN QN′ ⇒ + = + =e NN( ′) =e MM( ′) =2a
This means that for any point on the ellipse, the sum of its distances from the two foci is a constant, and equal to the major 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.
Q
F F′ QF QF′ F F′
The two definitions of the ellipse, one through ‘eccentricity’ and one through the ‘constant sum of distances’, are equivalent.
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 string of length L > l between these two pegs. Using your pen, stretch this string away from the pegs, so that it becomes taut :
l l
Fig - 7
Now always keeping the string taut, mark out a complete revolution on the paper with the pen. You’ve just made an ellipse ! This is because keeping the string taut ensures that the sum of the distances of the tip of your pen from the two foci is always constant, equal to L.
l
Move the pen while keeping the string taut
Fig - 8
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 form obtained earlier, we have
2ae=l 2a=L l e L ⇒ =
FOCAL DISTANCES: Let
2 2 2 2 1 x y
a +b = be the ellipse and F, F ' be the two foci (refer to fig - 6). The focal
OF Q(x, y) distances of Q are simply the two distances QF and QF . a QF eQN e x a ex e = = + = + a QF eQN e x a ex e ′= ′= − = −
Thus, the two focal distances of any point (x, y) are (a + ex) and (a – ex). The sum is 2a, as expected.
LATUS RECTUM : This is the chord passing through any of the two foci and perpendicular to the major axis. To evaluate the length of the latus-rectum, we must evaluate the y-coordinates of the two extremities of the latus rectum. In the equation of the ellipse
2 2 2 2 1 x y a +b =
we substitute x= ±ae ( the x-coordinates of the two foci). Thus
2 2 2 2 1 x y b a = − 2 2 2 (1 ) a e = − 2 (1 ) y= ±a −e 2 b a = ±
Thus, the length l of the latus-rectum is
2
2b
l a
=
Find the lengths of the major and minor axes and the foci for these ellipses:
(a) 2 2 16x +25y =400 (b) 2 2 4 2 0 x + y − x= (c) 2 2 25x +16y =400 Example – 1
Solution: (a) The given equation can be written in the form of the standard equation of an ellipse : 2 2 16 25 1 400 400 x y + = 2 2 1 25 16 x y ⇒ + = 2 2 2 2 1 5 4 x y ⇒ + =
Thus, comparing with the standard form, we have a = 5, b = 4
The major axis is of length 2a = 10 and the minor axis is of length 2b = 8. The eccentricity of the ellipse is
2 2 3 1 5 b e a = − = Thus, the two foci are at (±ae, 0), i.e., at ( 3, 0).±
(b) We first rearrange the equation of the ellipse to standard form :
2 2 4 2 1 x + y − x= 2 2 (x 1) 4y 1 ⇒ − + = 2 2 2 2 ( 1) 1 1 1 ( ) 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 is
( )
2 2 2 2 1 1 1 2 X + Y = Since a = 1 and 1, 2b= the major and minor axes are 2 and 1 respectively, while the eccentricity
is 1 1 3.
4 2
e= − = Thus, the two foci are at (in the X-Y system): 3 , 0 2 ±
In the original x-y system, we use the reverse transformation x→ +X 1, y→Y, so that the foci in the original system are at
3 1 , 0 2 ±
(c) 2 2 25x +16y =400 2 2 1 16 25 x 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 the x-axis but along the y-axis, and obviously, since the foci lie on the major axis, the foci will also lie on the y-axis. The major and minor axes are of lengths 2b = 10 and 2a = 8 respectively. The
eccentricity is 2 2 1 a e b
= − (a and b get interchanged ) 3. 5
= Thus, the two foci lie at (0,±be), i.e. at (0, ±3). Fig - 9 y (0 5), F1 F2 (4 0), (-4 0), (0 5) ,-x
Since the major-axis is along the -axis, the two foci also lie along the -axis. Their coordinates are
y
y
F = 1 (0, 3)
F = 2 (0, –3)
and
The length of the latus rectum in this case will be given by
2
2a
b and will be equal to
32 5
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 1 x y a +b = If a > b If a < 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) x a and x a e e = = − 2 2 1 b a − 2 2b a a ex± (0, b) and (0, -b) (0, be) and (0, -be) 2b (along y-axis) 2a (along x-axis) y b and y b e e = = − 2 2 1 a b − 2 2a b b±ey
And lastly, if the equation of the ellipse is 2 2 2 2 ( ) ( ) 1 x y a b − α + −β =
instead of the usual standard form, we can use the transformation X → − αx and Y → − βy (basically a translation of the axes so that the origin of the new system coincides with ( , ).α β The equation then becomes
2 2 2 2 1 X Y a +b =
We can now work on this form, use all the standard formulae that we’d like to and obtain whatever it is that we wish to obtain. The final result (in the x-y system) is obtained using the reverse transformation
x→ + αX and y→ + βY .
Find the equation of the ellipse with centre at the origin, foci at ( 1, 0)± and 1. 2 e=
Solution: Let the major axis be of length 2a. The distance between the two foci, 2ae, is equal to 2 for this example. 2ae=2 1 2 a e ⇒ = = Thus, 2 (1 ) 3 b=a −e =
The equation of the ellipse is therefore
2 2
1
4 3
x + y =
Find the equation of the ellipse whose major and minor axes lie along the lines x−3y+ =3 0 and 3x+ − =y 1 0 and whose lengths are 6 and 2 6 respectively.
Example – 2
Solution: The equation to the ellipse will obviously not be in the standard form since the axes are not along the coordinate axes. However, we can use the coordinate axes formed by these two lines as our reference frame Fig - 10 P X,Y ( ) B2 L x y2: 3 + – 1= 0 A A1 2= 6 A2 Y A1 X L x – y1: 3 + 3 = 0 B B1 = 2 62 O B1
Assume any point on the ellipse as
referred 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 original axes (not shown) and (X, Y) with respect to the new axes, the L1 – L2 system. In this new system, the equation of the ellipse is simply
2 2 2 2 1 X Y a +b =
where a = 3 and b= 6. Thus, the equation is
2 2
1
9 6
X +Y =
...(1)
Now, we wish to write the equation of the ellipse in the x-y coordinate system. For this purpose, we make the following observations from the figure:
What is X ? It is simply the perpendicular distance of P from L2:
3 1
10
x y X = + −
Similarly, Y is simply the perpendicular distance of P from L1 :
3 3
10
x y Y = − +
Thus, using (1), the equation of the ellipse in x-y form is
2 2 (3 1) ( 3 3) 1 90 60 x+ −y x− y+ + = 2 2 21x 6xy 29y 6x 58y 151 0 ⇒ − + + − − =
As an exercise find the centre and the foci of this ellipse.
(Hint : The centre is simply the intersection of L1 and L2. The foci are at (±ae, 0) 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 on
either 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) and an arbitrary fixed straight line lx+my+ =n 0 as the focus and directrix of an ellipse with eccentricity e, its equation 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 e x y = 2 2 2 2 2 2 ( ) (x h) (y k) e lx my n l m + + ⇒ − + − = +
This gives us the general equation of an ellipse with a given eccentricity, focus and directrix. However, we will almost always be using the (simple !) standard form of the equation.
Given an ellipse
2 2 2 2
( , ) :x y 1 0
S x y
a +b − = and a fixed point P x( ,1 y1), how will we determine whether P lies
inside or outside the ellipse ?
Solution: The approach we will follow here is the same as the one we used in circles and parabolas to solve the same question. Suppose P lies external to the ellipse as shown below. Drop a vertical line from P intersecting the ellipse at Q.
Fig - 11 y P x , y( 1 1)
Q x , y( 1 0)
x
Since Q lies on the ellipse, we have
2 2 0 1 2 2 1 y x a +b = 2 2 2 1 0 1 2 x y b a ⇒ = − Example – 4
Since P lies external to the ellipse, we must have y1 > y0 so that 2 2 1 0 y > y 2 2 2 1 1 1 2 x y b a ⇒ > − 2 2 1 1 2 2 1 0 x y a b ⇒ + − > 1 1 ( , ) 0 S x y ⇒ >
This condition must be satisfied if P lies outside the ellipse. (Convince yourself that wherever P may be outside the ellipse, this condition must hold). Similarly, P lies inside the ellipse if S x( ,1 y1)<0. We can 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 y S x y P x y S 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 metres while the distance between the flag posts is 8 meters. What is the area that the race track encloses ?
Solution: Fig - 12 y x P1 P2 AP + AP A 1 2is always
10 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 4
10 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 8 PP = ae= 5 a ⇒ = Thus, 2 (1 ) 3.
b=a −e = The equation of the elliptical race track is
2 2
1
25 9
x + 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 y a +b =
To find the area, we divide the ellipse into elemental strips of width dx : one such strip is shown below.
Fig - 13 y x x h x
xrepresents the height
of an elemental strip at distance from the centre. Note from the equation of the ellipse that
h = x
b a a – x
2 2
hx
The area of the elemental strip shown is 2 x
dA= h dx
2b a2 x dx2 a
= −
The area of the right half of the ellipse is therefore
2 2 half 0 2ba A a x dx a =
∫
− 2ab π = The total area of the ellipse is thushalf
2
A= ×A = πab
For the current example, the area becomes
5 3 15 sq. mt.
Show that the sum of the reciprocals of the squares of any two diameters of an ellipse which are at right angles to one another is a constant.
Solution: By a diameter of an ellipse, we mean any chord which passes through its centre.
Let
2 2 2 2 1 x y
a +b = be the ellipse and let AB and CD be any two diameters of the ellipse perpendicular
to each other. Fig - 14 y x C A B D 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 θ, than that of CD is obviously .
2
π +θ Thus, we get the co-ordinates of A, B, C and D as
1 1 1 , ( cos , sin ) A B≡ ± r θ r θ 2 2 , cos , sin 2 2 C D≡ ±r π+ θ r π+ θ = ± −( r2sin ,θ r2cos )θ These coordinates must satisfy the equation of the ellipse; we therefore obtain :
2 2 2 2 1 1 2 2 cos sin 1 r r a b θ+ θ = and 2 2 2 2 2 2 2 2 sin cos 1 r r a b θ+ θ = 2 2 2 2 2 1 1 cos sin r a b θ θ ⇒ = + and 2 22 22 2 1 sin cos r a b θ θ = +
Adding these two relations, we have
2 2 2 2 1 2 1 1 1 1 r +r = a +b 2 2 2 2 1 2 1 1 1 1 (2 )r (2 )r 4a 4b ⇒ + = + 2 2 2 2 1 1 1 1 4 4 AB CD a b ⇒ + = + (a constant)
This proves the assertion stated in the question.
Let
2 2 2 2 1 x y
a +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 PQ will always be divided in a fixed ratio by the ellipse.
Solution: Fig - 15 y x P O R Q θ P = a ( cosθ, a sin )θ R = a ( cosθ, b sin )θ Q = a ( cosθ, 0) The justification for these coordinates is given below.
As is evident, the radius of this circle, called the auxiliary circle of the ellipse, is a, so that its equation is
2 2 2 x +y =a
Now any point P on this circle can be taken in parametric form as P≡( cos , sin )a θ a θ where θ is the angle that OP makes with the horizontal.
To evaluate the y-coordinate of R, we substitute x=acosθ in the equation of the ellipse :
2 2 2 2 2 cos 1 a y a θ + =b sin y b ⇒ = θ
Thus, R is the point ( cos , sin )a θ b θ while Q is simply ( cos , 0).a θ We now see that
sin sin 1 sin PR a b a RQ b b θ − θ = = − θ which is independent of θ, proving the stated assertion.
There is one significant fact that we can learn about the ellipse
2 2 2 2 1. x y
a +b = We obtained the
coordinates of R as ( cos , sin ).a θ b θ This tells us that an alternative way to specify an ellipse is in terms of a parameter θ:
cos , sin
x=a θ y=b θ
This is referred to as the parametric form of the ellipse
2 2 2 2 1. x y
a +b = θ is called the eccentric angle
of the point ( , )x y ≡( cos ,a θ bsin ).θ It is important to note that θ is not the angle that ( cos ,a θ bsin )θ makes with the horizontal; it is the angle which the corresponding point on the auxiliary circle makes with the horizontal. Depending on what value θ takes in the range [0, 2 )π
(or (- , ]),π π the parametric form ( cos , sin )a θ b θ gives us different points on the circumference of the ellipse.
The point ( cos , sin )a θ b θ in sometimes simply referred to as the point θ.
What are the eccentric angles of the extremities of the latus-recta in the ellipse
2 2 2 2 1 ? x y a +b =
Solution: As discussed earlier, the co-ordinates of the end-points of the latus-recta are
2 , b . ae a ± ± Thus, if we assume the required eccentric angle to be φ, we have
cos a φ = ±ae 2 sin b b a φ = ± This gives us four values of φ, given by
tan b
ae φ = ±
corresponding to the four extremities of the two latus-recta :
Fig - 16 y x A' φ
The two latus-recta and meet the auxiliary circle in ', ', C' and '. The slopes of the lines joining 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 CD A B D A φ C' D' B' A B D C Example – 8
EQUATION OF : Consider two points θ and φ lying on the ellipse
2 2 2 2 1. x y a +b =
A CHORD We wish to determine the equation of the chord joining these two points. JOINING θ and φ Fig - 17 y x θ( cosa θ, b sin )θ φ( cosa φ, b sin )φ
Using the two point form, we have
sin sin sin
cos cos cos
y b b b x a a a − θ = θ − φ − θ θ − φ cos sin 2 cos sin 2 y b b x a a θ + φ − θ − ⇒ = θ + φ − θ
cos sin cos cos sin sin
2 2 2 2 x y a b θ + φ θ + φ θ + φ θ + φ ⇒ + = θ + θ
cos sin cos
2 2 2 x y a b θ + φ θ + φ θ − φ ⇒ + =
This is the most general equation of a chord joining any two arbitrary points θ and φ 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 θ1 and θ2 on the ellipse
2 2 2 2 1 x y
a +b = intersects the major-axis in (h, 0).
Show that 1 2 tan tan . 2 2 h a h a θ θ − = + Example – 9
Solution: By the result we just derived, the equation of the chord joining θ1 and θ2 is
1 2 1 2 1 2
cos sin cos
2 2 2 x y a b θ + θ θ + θ θ − θ + =
If this passes through (h, 0), we have
1 2 1 2 cos cos 2 2 h a − θ + θ = θ − θ 1 2 1 2 cos 2 cos 2 h a θ − θ ⇒ = θ + θ 1 2 1 2 1 2 1 2 cos cos 2 2 cos cos 2 2 h a h a θ − θ θ + θ − − ⇒ = θ − θ θ + θ + + tan 1 tan 2 2 2 θ θ =
A circle intersects the ellipse
2 2 2 2 1 x y
a +b = in four points A, B, C and D whose eccentric angles are respectively 1, 2, 3
θ θ θ and θ4. Show that θ + θ + θ + θ1 2 3 4 will be an integral multiple of 2 .π
Solution: Suppose that the circle cuts the ellipse as shown :
Fig - 18 y x A( )θ1 B( )θ2 C( )θ3 D( )θ4 Example – 10
Using the general equation of a chord joining two arbitrary points on an ellipse derived earlier, we can write L1 =0 and L2 =0, the equation of AB and CD respectively. Doing this has the advantage that we can now write (using a family of circles approach) any circle passing through the four point A, B, C and D as
1 2 0 S+ λL L =
where S = 0 is the equation of the ellipse. Imposing the necessary condition of λ for this equation to represent 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 4 1 2 1 2 2 2 2 2 1 2 3 4 Equation of Equation of cos sin cos sin 2 2 2 2 1 0 cos cos 2 2 AB CD x y x y a b a b x y a b θ + θ θ + θ θ + θ + θ + θ + + − + λ = θ − θ θ − θ − − !""""""#""""""$ !"""""""#"""""""$ This represents a circle if
2 2 Coeff. of x =Coeff. of y Coeff. of xy=0 3 4 3 4 1 2 1 2 2 2 2 2 1 1
cos cos sin sin
2 2 2 2 a a b b θ + θ θ + θ λ θ + θ λ θ + θ ⇒ + = +
and cos 1 2 sin 3 4 cos 3 4 sin 1 2 0
2 2 2 2 θ + θ θ + θ θ + θ θ + θ + =
The second relation gives
1 2 3 4 cos 0 2 θ + θ + θ + θ = 1 2 3 4 2 n θ + θ + θ + θ ⇒ = π , n∈% 1 2 3 4 2n ⇒ θ + θ + θ + θ = π , 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 be advantageous to remember it.
If px+qy+ =r 0 cuts the ellipse
2 2 2 2 1 x y
a +b = in points whose eccentric angles differ by 2,
π
show that
2 2 2 2 2
2
a p +b q = r
Solution: The two points of intersection can be assumed to be
( cos , sin ) P≡ a θ b θ cos , sin 2 2 Q≡a θ +π b θ +π ≡ −( asin , cos )θ b θ
Since P and Q both lie on the line px+qy+ =r 0, their coordinates satisfy its equation. We thus have
cos sin 0
ap θ +bq θ + =r and −apsinθ +bqcosθ + =r 0
cos sin
ap bq r
⇒ θ + θ = − and −apsinθ +bqcosθ = −r.
Squaring 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 a
a +b = < < Let the line parallel to the y-axis passing through P meet
the circle 2 2 2
x +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.
Solution: The circle 2 2 2
x +y =a is the auxiliary circle of the given ellipse. y x Q a ( cosθ, a sin )θ P a ( cosθ, b sin )θ Fig - 19 R Example – 11 Example – 12
We can assume point P to be ( cos , sin )a θ b θ so that Q will be ( cos , sin ).a θ a θ The point R(h, k) divides the segment PQ internally in the ratio r : s. Thus,
cos cos sin sin
, ar as ar bs h k r s r s θ + θ θ + θ = = + +
We need to eliminate θ from these two relations to obtain a relation between h and k. Thus,
( ) cos h, sin k r s a ar bs + θ = θ = + Squaring and adding the two gives us the required relation :
2 2 2 2 2 ( ) 1 ( ) h r s k a ar bs + + = + The required locus of R is
2 2 2 2 2 ( ) 1 ( ) x r s y a ar bs + + = +
which is an ellipse, as might have been expected.
Consider the ellipse
2 2 2 2 1. x y
a +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 φ, while ∠SFP= θ. Prove that 1 tan tan 2 1 2 e e θ= + φ − Solution: Fig - 20 y x P a ( cosφ, b sin )φ T θ F ae,( 0) S a,( 0) Drop a perpendicular from onto the major axis. The point is then
( cos 0)
PT P
T
T = a , φ
We have, from the figure,
cos FT
PF − θ = Example – 13
Since F is the focus, note that PF will simply be e times the distance of P from the directrix x a. e = Thus, PF e a acos e = − φ = −a aecosφ (cos ) cos cos (1 cos ) 1 cos a e e a e e φ − φ − ⇒ θ = = − φ − φ ...(1) Thus, 2 1 cos (1 )(1 cos ) tan 2 1 cos (1 )(1 cos ) e e θ − θ + − φ = = + θ − + φ {Using (1)} (1 )tan2 (1 ) 2 e e + φ = − 1 tan tan 2 1 2 e e θ + φ ⇒ = −
Consider a focal chord AB of the ellipse
2 2 2 2 1 x y
a +b = where the eccentric angles of A and B are θ1 and θ2
respectively. If e is the eccentricity of the ellipse, prove that
1 2 1 2 sin sin sin( ) e= θ + θ θ + θ Solution: The chord joining A and B has the equation
1 2 1 2 1 2
cos sin cos
2 2 2 x y a a θ + θ θ + θ θ − θ + = ...(1)
Since AB is a focal chord of the ellipse (say, it passes through F ae1( , 0)), the coordinates of the focus must satisfy (1) so that we have
1 2 1 2 cos cos 2 2 e θ + θ = θ − θ 1 2 1 2 cos 2 cos 2 e θ − θ = θ + θ ...(2) Example – 14
Multiplying the numerator and denominator of the RHS of (2) by 1 2 2 sin , 2 θ + θ we have 1 2 1 2 1 2 1 2 2 cos sin 2 2 2 cos sin 2 2 e θ − θ θ + θ = θ + θ θ + θ 1 2 1 2 sin sin sin( ) θ + θ = θ + θ Consider an ellipse 2 2 2 2 1 x y
a +b = and a variable line y=mx c+ . What is the condition on m and c such that the
line
(a) intersects the ellipse in two distinct points ? (b) touches the ellipse ?
(c) does not intersect with the ellipse ?
Solution: As we’ve done in the case of circles and parabolas, to find the intersection (points) of the line and the ellipse, we must solve their equations simultaneously;
2 2 2 2 1 ; x y y mx c a +b = = + 2 2 2 2 ( ) 1 x mx c a b + ⇒ + = 2 2 2 2 2 2 2 2 (a m b x) 2a mcx a c( b ) 0 ⇒ + + + − = ...(1) The line y=mx c+
(a) intersects the ellipse (b) touches the ellipse
(c) 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 be
2 2 2 2
c =a m +b
Thus, we can say that the line 2 2 2
y=mx± a m +b will always be a tangent to the ellipse, whatever
may be the value of m. We discuss tangents in more detail in the next section.
Q. 1 Let P be a variable point on the ellipse
2 2 2 2 1 x y
a +b = with foci S1 and S2. Find max(area(∆PS S1 2)).
Q. 2 Find the equation of the ellipse with foci at (0, ±4) and eccentricity 4. 5
Q. 3 Show that 2 2
4 2 16 13 0
x + y + x+ y+ = is the equation of an ellipse. Where are its foci ?
Q. 4 Find the equation of the ellipse whose foci are ( 2, 3)± and whose semi-minor axis is of length 5. Q. 5 A straight rod of length l slides between the x-axis and the y-axis, as shown. Show that the locus of
its mid-point is an ellipse. What is its eccentricity ? y
x l
Q. 6 Show that the traingle with vertices (1, 2), (3, –1) and (–2, 1)lies completely inside the ellipse
2 2
2 13.
x + y =
As in circles and parabolas, the equation of a tangent to a given ellipse can take various different forms, all of which we discuss in this section. We will use the ellipse
2 2 2 2 1 x y
a +b = as our standard throughout this discussion.
TANGENTS AT : Consider the ellipse
P(x1, y1) 2 2 2 2 ( , ) : x y 1 S x y a +b =
and a point P(x1, y1) lying on this ellipse. Thus, 2 2 1 1 2 2 1 x y a +b = ...(1)
The slope mT of the tangent at P(x1, y1) can be obtained by evaluating the derivative of the curve at P. For this purpose, we differentiate the equation of the ellipse :
2 2 2 2 0 x y dy a + b dx = 2 2 dy b x dx a y − ⇒ = 1 1 2 1 2 ( , ) 1 T P x y dy b x m dx a y ⇒ = = −
The equation of the tangent can now be obtained using point-slope form :
2 1 1 2 1 1 ( ) b x y y x x a y − − = − 2 2 1 1 1 1 2 2 2 2 xx yy x y a b a b ⇒ + = + ...(2)
Using (1), the RHS in (2) is 1 so that the equation of the ellipse is
1 1 2 2 1 xx yy
a + b =
The equation obtained for the tangent can be, as in the case of circles and parabolas, written concisely in the form
1 1
( , ) 0
T x y =
TANGENT AT P(θ θ θ θ θ ): If the point P is specified in parametric form instead of cartesian form, we simply substitute
1 cos , 1 sin
x →a θ y →b θ in the equation of the tangent obtained above. Thus, the equation in this case is
cos sin 1
x y
a θ +b θ =
TANGENT OF : In example-15, we proved that any line of the form
SLOPE m 2 2 2
y=mx± a m +b
is a tangent to the ellipse
2 2 2 2 1, x y
a +b = whatever the value of m may be.
As an exercise, show that this tangent touches the ellipse at the point
2 2 2 2 2 , 2 2 2 a m b a m b a m b ± + + ∓
Also show that from any point P, in general two tangents (real or imaginary)can be drawn to the ellipse (use the approach followed in Circles)
Tangents drawn at A( )θ1 and B(θ2) on the ellipse
2 2 2 2 1 x y
a +b = intersect in P. Find the coordinates of P.
Solution: The equations of the tangents at A and B, using the parametric form for the tangent, are
1 1 cos sin 1 x y a θ +b θ = ...(1) 2 2 cos sin 1 x y a θ + b θ = ...(2) 2 1
(1) cos× θ −(2) cos× θ gives
1 2 2 1
sin( ) cos cos
y b θ − θ = θ − θ 2 1 1 2 (cos cos ) sin( ) y b θ − θ ⇒ = θ − θ 1 2 1 2 sin 2 cos 2 b θ + θ = θ − θ Example – 16
Similarly,
2 1
(1) sin× θ −(2) sin× θ gives
1 2 1 2 cos 2 cos 2 a x θ + θ = θ − θ
Thus, the coordinates of the point of intersection P are
1 2 1 2 1 2 1 2 cos sin 2 2 , cos cos 2 2 a b P θ + θ θ + θ ≡ θ − θ θ − θ
Find the locus of a moving point P such that the two tangents drawn from it to an ellipse are perpendicular.
Solution: Let the equation of the ellipse be
2 2 2 2 1 x y
a +b = and P be the point (h, k).
Any tangent to this ellipse is of the form
2 2 2 y=mx+ a m +b
If this passes through P(h, k), we have
2 2 2 k=mh+ a m +b 2 2 2 2 (k mh) a m b ⇒ − = + 2 2 2 2 2 (h a m) 2hkm k b 0 ⇒ − − + − =
As expected, we obtain a quadratic in m which will give us two roots, say m1 and m2. Since the tangents through P are perpendicular, we have
1 2 1 m m = 2 2 2 2 1 k b h a − ⇒ = − − 2 2 2 2 h k a b ⇒ + = +
Thus, the locus of P is
2 2 2 2 x +y =a +b
which is a circle and is termed the Director Circle of the ellipse. From any point on the Director Circle of an ellipse, the two tangents drawn to the ellipse are perpendicular.
y
x
From any point on the director circle of an ellipse, the two tangents drawn to the ellipse are perpendicular.
Fig - 21 P
Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse
2 2 2 2 1 x y a +b = is 2 . b Solution: We can assume an arbitrary tangent to this ellipse to be
2 2 2
y=mx+ a m +b ...(1)
The perpendicular distances of the two foci, F ae1( , 0) and F2(−ae, 0) from the line given by (1) are
2 2 2 1 2 1 mae a m b d m + + = + 2 2 2 2 2 1 mae a m b d m − + + = + We thus have, 2 2 2 2 2 2 1 2 2 1 a m b a m e d d m + − = + 2 2 2 2 2 (1 ) 1 a m e b m − + = + 2 2 2 2 1 m b b m + = + 2 b = Example – 18
Find the locus of the foot of the perpendicular drawn from the origin upon any tangent to the ellipse
2 2 2 2 1. x y a +b =
Solution: We can assume an arbitrary tangent to this ellipse as
2 2 2 y=mx+ a m +b
Let the foot of perpendicular from the origin upon this tangent be P(h, k):
Fig - 22 y P h,k( ) S O x Thus, 2 2 2 k=mh+ a m +b ...(1)
Also,OP⊥SP ( the tangent)
1 k m h ⇒ × = − h m k ⇒ = − ...(2)
Using (2) in (1), we obtain a relation between h and k as
2 2 2 2 2 2 2
(h +k ) =a h +b k
Thus, the required locus of P is
2 2 2 2 2 2 2
(x +y ) =a x +b y
Prove that the portion of the tangent to any ellipse intercepted between the curve and a directrix subtends a right angle at the corresponding focus.
Example – 19
Solution: The following diagram makes the phrase “corresponding focus” clear : Fig - 23 y P x F1 T1 F2 T2
We need to prove that ∠PF1T = 1 ∠PF T2 2 =
π
2
Let the equation of the ellipse be
2 2 2 2 1 x y
a +b = and assume P to be ( cos , sin ).a θ b θ F1 and F2 are (–ae, 0) and (ae, 0) respectively.
The equation of the tangent at P is
cos sin 1
x y
a θ +b θ =
T1 and T2 can now be evaluated since we know their x-coordinates as a e − and a e respectively. 1 ( cos ) : sin a b e T x y e e − + θ = ⇒ = θ 1 ( cos ) , sin a b e T e e − + θ ⇒ ≡ θ Similarly, 2 ( cos ) , sin a b e T e e − θ ≡ θ Now we evaluate the appropriate slopes :
1 1 2 ( cos ) ( cos ) sin sin (1 ) F T b e b e e m a a e ae e + θ − + θ θ = = θ − − 1 sin sin cos ( cos ) PF b b m a ae a e θ θ = = θ + + θ 1 1 1 2 2 2 2 2 1 (1 ) F T PF b b m m a e b − − ⇒ × = = = − − which implies FT1 1 ⊥PF1
Similarly, we can show that F T2 2 ⊥PF2.
Common tangents are drawn to the parabola 2
4
y = x and the ellipse
2 2
1
16 6
x + y =
touching the parabola at A and B and the ellipse at C and D. What is the area of the quadrilateral ABDC ?
Solution: Fig - 24 y O x A C E P D y = x2 4 Q B An approximate figure showing the common tangents and intersecting in (which will lie on the axis due to the symmetry of the problem)
AC BD
E
Any tangent to the parabola 2
4
y = x can be written in the form 1
y mx
m
= +
This line touches the ellipse if the condition for tangency, 2 2 2 2
c =a m +b is satisfied, i.e. if 2 2 1 10m 6 m = + giving 1 2 2 m= ±
Thus, the two tangents AE and BE are
1
2 2 2 2
y= ± x+
which evidently intersect at E( 4, 0).−
The point of contact for the parabola 2 4 y = ax is given by 2 2 , . a a m m
Thus A and B have the coordinates (8, 4 2)± so that
8 2
AB=
The point of contact for the ellipse will be
2 2 2 2 2 , 2 2 2 . a m b a m b a m b ± + + ∓
Thus, C and D will have the coordinates 2, 3 2 − ± so that 3 2 CD=
Finally, PQ can now easily be seen to be 8 + 2 = 10. The area of quadrilateral ABDC (which is actually a trapezium) is 1 ( ) 2 AB CD PQ ∆ = × + × =55 2 sq units
Prove that in any ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to point of contact meet on the corresponding directrix.
Solution: Let the ellipse be
2 2 2 2 1 x y
a +b = and let a tangent be drawn to it at an arbitrary point P a( cos , sin )θ b θ as
shown : Fig - 25 y x O P F T Q
We need to show that the perpendicular from F onto this tangent, i.e., FT, and the line joining the centre to the point of contact, i.e. OP intersect on the corresponding directrix; in other words, we need to show that the x-coordinate of Q as in the figure above is x a.
e = Example – 22
The equation of the tangent at P is xcos ysin 1.
a θ +b θ =
The slope of this tangent is T cot b m
a −
= θ
Therefore, the slope of FT is
tan FT a m b = θ The equation of FT is : atan ( ) FT y x ae b = θ − ...(1)
The equation of OP is simply
: btan OP y x a = θ ...(2)
Comparing (1) and (2) gives x a ,
e
= which proves the stated assertion.
Find the coordinates of all the points on the ellipse
2 2 2 2 1 x y
a +b = for which the area of ∆PON is the maximum
where O is the origin and N is the foot of the perpendicular from O to the tangent at P.
Solution: We can assume the point P to be ( cos , sin )a θ b θ so that the tangent at P has the equation
cos sin 1 x y a θ +b θ = ...(1) Fig - 26 O P a , ( cosθ b sin )θ N y x
To evaluate the area of ∆PON, we first need the coordinates of the point N. The equation of ON is
0 atan ( 0) y x b − = θ − sin cos 0 ax by ⇒ θ − θ = ...(2) Example – 23
The intersection of (1) and (2) gives us the coordinates of N as
2 2
2 2 2 2 2 2 2 2
cos sin
,
sin cos sin cos
ab a b N a b a b θ θ ≡ θ + θ θ + θ
The length PN can now be evaluated using the distance formula :
2 2 2 2 2 2 ( ) sin cos sin cos a b PN a b − θ θ = θ + θ
The length ON is simply the perpendicular distance of O from the tangent at P given by (1) :
2 2 2 2 sin cos ab ON a b = θ + θ
Thus, the area of ∆OPN is
1 2 PN ON ∆ = × × 2 2 2 2 2 2 ( ) sin cos 2 sin cos ab a b a b − θ θ = θ + θ 2 2 1 2 tan cot a b a b b a − = θ + θ
The expression atan bcot
b θ +a θ is of the form 1
y y
+ whose minimum magnitude is 2, when
tan 1 a b θ = ± tan b a ⇒ θ = ± 2 2 2 2 sin b , cos a a b a b ⇒ θ = ± θ = ± + +
When this minimum is achieved, ∆ is maximum. Thus, the possible coordinates of P for which area (∆OPN) is maximum are
2 2 2 2 , 2 2 a b P a b a b ≡ ± ± + +
A straight line L touches the ellipse
2 2 2 2 1 x y
a +b = and the circle
2 2 2
x +y =r (where b < r < a). A focal chord of the ellipse parallel to L meets the circle in A and B. Find the length of AB.
Solution: L is a common tangent to the ellipse and the circle. We can assume the equation of L to be (using the form of an arbitrary tangent to an ellipse) :
2 2 2 y=mx+ a m +b Fig - 27 x y A B F1 N L O
Since L is a tangent to the circle too, its distance from (0, 0) must equal r. Thus,
2 2 2 2 1 a m b r m + = + 2 2 2 2 r b m a r − ⇒ = ± − ...(1)
The equation of AB( which passes through F ae1( , 0)) can now be written as
0 ( )
y− =m x−ae
mx y mae
⇒ − = ...(2)
To evaluate the length AB, one alternative is to find the intercept that the circle 2 2 2
x +y =r makes on the line AB given by (2).
However, a speedier approach would be to use the Pythagoras theorem in ∆OAN. OA is simply the radius r. ON is the perpendicular distance of O from the line given by (2). Thus,
2 1 mae ON m = + Example – 24
Finally, we have 2 AB= AN 2 2 2 OA ON = − 2 2 2 2 2 2 1 m a e r m = − + 2 2 2 2 2 2 2 2 2 2 2 ( ) 2 ( ) r b a e a r r a b a b − − = − − − = 2 b
The length of the chord AB is equal to 2b, the same as the minor axis of the ellipse.
Find the angle of intersection of the ellipse
2 2 2 1 x y a b
2
+ = and the circle 2 2
.
x +y =ab
Solution: The semi-major and semi-minor axis of the ellipse are of lengths a and b respectively whereas the radius of the circle is ab. Note that
b< ab <a
Thus, the circle will intersect (symmetrically) the ellipse in four points.
Fig - 28 x y P We need to find φ Example – 25
Consider any point of intersection, say P, the one in the first quadrant. The coordinates of P can be assumed to be ( cos , sin ).a θ b θ Since P also lies on the circle, we have
2 2 2 2 cos sin a θ +b θ =ab 2 2 2 (a b ) cos b a b( ) ⇒ − θ = − 2 2 cos sin b a b a a b ⇒ θ = + ⇒ θ = + ...(1)
At P, the tangent to the ellipse has the slope
3 2 cot E b b m a a − = θ = − (Using (1))
while the tangent to the circle has the slope
1 2 cot C a a m b b − = θ = − (Again, using (1)) Thus, the angle of intersection is given by
3 1 2 2 tan 1 1 C E C E a b m m b a a b b m m ab a − − + φ = = = + + 1 tan a b ab − + ⇒ φ =
Q. 1 Prove that the tangents at the extremities of the latus rectum of an ellipse intersect on the corresponding directrix.
Q. 2 Let P and Q be points on the ellipse
2 2 2 2 1 x y
a +b = whose eccentric angles differ by 2.
π
Tangents at P and Q intersect at R. What is the locus of R ?
Q. 3 Prove that the locus of the mid-point of the portion of the tangent to the ellipse
2 2 2 2 1 x y
a +b = intercepted
between the coordinate axes is 2 2 2 2 2 2
4 .
b x +a y = x y
Q. 4 We generalise Question - 2 (above) in this question. What is the locus of the point of intersection of
tangents to the ellipse
2 2 2 2 1 x y
a +b = at points whose eccentric angles differ by a constant 2α.
Q. 5 Find the locus of the foot of perpendicular on any tangent to the ellipse
2 2 2 2 1 x y
a +b = from either of its
foci.
Q. 6 A family of ellipses have the same major axis, but different minor axis. Prove that the tangents at the end-points of their latus - rectums will always pass through a fixed point.
Q. 7 Let P be any point on the ellipse
2 2 2 2 1 x y
a +b = with y-coordinate k. Prove that the angle between the
tangent at P and the focal chord through P is
2 1 tan b . aek − Q. 8 A tangent to 2 2 2 2 1 x y
a +b = cuts the axes in A and B and touches the ellipse at P in the first quadrant.
What is the equation of this tangent if AP = PB ?
Q. 9 If the tangent at any point on
2 2 2 2 1 x y
a +b = makes an angle α with the major axis and an angle β with
the focal radius of the point of contact, show that the eccentricity of the ellipse e satisfies cos . cos
e= β
α Q. 10 Prove that the tangent at any point on the ellipse bisects the external angle between the focal radii of
that point.
We will find, in this section, that the equations of chord of contact, chord bisected at a given point, and pair of tangents from a point take the same form as they do in the case of circles and parabolas. We will use the ellipse
2 2 2 2 ( , ) x y 1 0 S x y a b ≡ + − = in this discussion.
CHORD OF : Two tangents are drawn from an external point P(x1, y1) to the ellipse
2 2 2 2 1, x y a +b =
CONTACT FROM touching this ellipse at points A and B. We need to find the equation of AB, the chord of
P(x1, y1): contact. Fig - 28 y P x ,y( 1 1) A x B AB P is the chord of contact for the tangents drawn from to the ellipse
To determine this equation, we’ll follow precisely the same approach as we did in the case of circles and parabolas.
Assume the coordinates of A and B to be (x2, y2) and (x3, y3) respectively. The tangents at A and B then have the equations :
2 2 2 2 1 xx yy a + b = and 3 3 2 2 1 xx yy a + b =
Since these tangents intersect at P x y, ( ,1 1) must satisfy both these equations. Thus,
1 2 1 2 2 2 1 x x y y a + b = and 1 3 1 3 2 2 1 x x y y a + b =
Now, from these two equations, we can deduce that ( ,x y2 2) and ( ,x y3 3) are actually
two solutions of the linear equation
1 1 2 2 1
xx yy
a + b =
which means that this must be the chord of contact.
The equation of the chord of contact can be written concisely as
1 1
( , ) 0
T x y =
CHORD BISECTED : Let the required chord be AB, where the coordinates of A and B are (x2, y2) and (x3, y3) AT P(x1, y1) respectively. Fig - 30 y x A x y( , )2 2 P x y( , )1 1 B x y( , )3 3
Since A and B lie on the ellipse, we have
2 2 2 2 2 2 1 x y a +b = ...(1) 2 2 3 3 2 2 1 x y a +b = ...(2) By (1) – (2), we have 2 3 2 3 2 3 2 3 2 2 ( )( ) ( )( ) 0 x x x x y y y y a b − + + + − = ...(3) We also have to use the fact that P is the mid-point of AB, so that
2 3 2 1 x + =x x and y2+ =y3 2y1 ...(4) Using (4) in (3), we obtain 1 2 3 1 2 3 2 2 2 ( ) 2 ( ) 0 x x x y y y a b − + − = 2 2 3 1 2 2 3 1 y y b x x x a y − ⇒ = − − ...(5)
Observe carefully that what (5) gives is simply the slope of AB.
Once we know the slope of AB, we can use the point-slope form to write its equation as : 2 1 1 2 1 1 ( ) b x y y x x a y − = − − 2 2 1 1 1 1 2 2 2 2 xx yy x y a b a b ⇒ + = +
If we subtract 1 from both sides, we have
2 2
1 1 1 1
2 2 1 2 2 1 xx yy x y
a + b − = a +b −
which can be written concisely as
1 1 1 1
( , ) ( , )
PAIR OF TANGENTS: Let Q(h, k) be any point on the pair of tangents drawn from P to the ellipse. FROM P(x1, y1): Fig - 31 y x P x y( , )1 1 Q h k( , )
The equation of PQ, by two-point form is
1 1 1 1 y y k y x x h x − = − − − 1 1 1 1 1 k y hy kx y x h x h x − − ⇒ = + − − ...(1)
PQ being a tangent to the ellipse, the condition 2 2 2 2
c =a m +b for tangency must be satisfied in (1). Thus, 2 2 2 2 1 1 1 1 1 hy kx k y a b h x h x − = − + − − ...(2)
As expected, (2) is a second degree equation in h and k, since the point Q(h, k) can lie on essentially two different lines (the two tangents). Thus, what (2) represents is the pair of tangents from P to the ellipse.
After some manipulation, the relation in (2) can be written as
2 2 2 2 2 1 1 1 1 2 2 1 2 2 1 2 2 1 h k x y hx ky a b a b a b + − + − = + −
Using (x, y) instead of (h, k), the equation to the pair of tangents becomes
2 2 2 2 2 1 1 1 1 2 2 1 2 2 1 2 2 1 x y xx yy x y a b a b a b + − + − = + −
which can be written concisely as
2 1 2 1 1 ( , ) ( , ) ( , ) T x y =S x y S x y or simply 2 1. T =SS
Find the locus of the point such that the chord of contact of the tangents drawn from it to the ellipse
2 2 2 2 1 x y a +b =
touches the circle 2 2 2
.
x +y =r
Solution: Let P h k( , ) be such a point.
Fig - 32 y
x
P B
A
The equation of the chord of contact AB from P(h, k) is
( , ) 0 T h k = 2 2 1 hx ky a b ⇒ + = 2 2 2 b h b y x a k k − ⇒ = + This is a tangent to the circle 2 2 2
,
x +y =r if the condition for tangency for the case of circles
2 2 2 (c =a (1+m )) is satisfied : 4 4 2 2 2 1 4 2 b b h r k a k = + 4 4 4 2 2 4 2 2 a b a r k b r h ⇒ = +
( ) ( )
2 2 2 2 2 2 1 h k a b r r ⇒ + =Thus, the locus of P is
( ) ( )
2 2 2 2 2 2 1 x y a b r r + = which is an ellipse Example – 26A variable chord AB of the ellipse
2 2 2 2 1 x y
a +b = subtends a right angle at its centre. Tangents drawn at A and B
intersect at P. Find the locus of P. Solution: Let P be the point (h, k).
Fig - 33 y x O A B P h,k( )
Since AB is the chord of contact for the tangents drawn form P, the equation of AB will be ( , ) 0 T h k = 2 2 1 hx ky a b ⇒ + = ...(1)
We can now write the joint equation of OA and OB by homogenizing the equation of the ellipse using the equation of the chord AB obtained in (1) :
Joint equation of AB : 2 2 2 2 2 2 2 x y hx ky a b a b + = + ...(2)
Since OA and OB are perpendicular, we must have
2 2 Coeff. of x +Coeff. of y =0 in (2) 2 2 2 4 2 4 1 1 0 h k a a b b ⇒ − + − = 2 2 4 4 2 2 1 1 h k a b a b ⇒ + = +
The locus of P is therefore
2 2 4 4 2 2 1 1 x y a +b =a +b Example – 27
A tangent drawn to the ellipse
2 2 1: 2 2 1
x y E
a +b = intersects the ellipse
2 2 2:
x y
E a b
a + b = + at P and Q. Prove that
the tangents drawn to E2 at P and Q intersect at right angles.
Solution: Let the point of intersection be R h k( , ).
Fig - 34 y x E1 P R h,k( ) E2 Q
PQ is the chord of contact for the tangents drawn from R h k( , ) to E2. Thus, the equation of PQ is
( , ) 0 T h k = hx ky a b a b ⇒ + = + ( ) bh b y x a b ak k − ⇒ = + +
PQ touches the inner ellipse E1 if the condition for tangency for ellipses 2 2 2 2
(c =a m +b ) is satisfied. Thus, 2 2 2 2 2 2 2 ( ) 2 2 b b h a b a b k a k + = + 2 2 2 ( ) h k a b ⇒ + = + ...(1)
By Example -17, any point ( ,h k′ ′) lying on the director circle of E2 must satisfy
2 2 ( ) ( ) h′ +k′ =a a b+ +b a b+ 2 (a b) = + ... (2)
From (1) and (2), it is evident that the point R h k( , ) itself lies on the director circle of E2, and thus, by definition of a director circle, ∠PRQ= °90 . Example – 28
