Download CHAPTER 14 ELLIPSE Advanced Exercise Page 203

CHAPTER 14 ELLIPSE
Advanced Exercise Page 203-204
1.
LHS ( PF1  PF2 ) 2  ((a  ex1 )  (a  ex1 )) 2  4e 2 x12
If P is ( x1 , y1 ) then equation of tangent at P must be
 d
Also
1
x12 y12

a 4 b4
1 x12 y12


d 2 a 4 b4

x12 y12

1
a 2 b2
xx1 yy1

1
a 2 b2
(*)
(**)
On eliminating y12 between (*) and (**), we get
x12  1 1  1
1 x12 1  x12 

 1 
  2  2  2  2
d 2 a 4 b2  a 2 
a a b  b


x12e2
b2 x12  b2 
b2 x12  b2  x12 a 2  b 2


.



1

1
1


1





a2
a2
d 2 a2  a2 
d 2 a2  a2  a2
 b2 
4a 2 1  2   4e2 x12
 d 

RHS  LHS
2.
SEE SOLVED EXAMPLE 7
3.
If P be (a cos  , b sin  ) then equation of tangent at P is
x
y
cos   sin   1
a
b
(i)
Also equation of normal at P is
Now height of OPN

1
cos 2  sin 2 
 2
a2
b
ax
by

 a 2  b2
cos  sin 
  distance O from (i)

ab
a 2 sin 2   b2 cos 2 
Base   distance of O from (ii)
1
and
(ii)

a 2  b2

a2
b2

cos 2  sin 2 
(a 2  b2 )sin  cos 
a 2 sin 2   b2 cos 2 
 Area A 
1 ab.(a 2  b 2 )sin  cos 
2 a 2 sin 2   b2 cos 2 
1
1
ab(a 2  b 2 )
ab(a 2  b 2 )
2
A  22

2
2
a tan   b cot 
a tan   b cot   2ab


max A 

1 ab.(a 2  b2 ) a 2  b 2

2
2ab
4
This will be attained when a tan   b cot 
 cos  
a
a b
2
2
 a2
 P is 
,
2
2
 a b
4.
,sin  
or tan  
b
a
b
a  b2
2

 ( with several equivalent points )
a 2  b2 
b2
Without loss of generality let us take A and B in the upper half plane and C in the lower plane.
Then it can be easily observed that if A be (a cos  , a sin  ) then B and C are the points

2

 a cos    3


2  

4



 , a sin   
  and  a cos   
3 
3




4  


 , a sin   
 respectively. (*)
3  


x2 y 2

 1 of the point
a 2 b2
P, Q, R respectively therefore P, Q, R must be the points (a cos  , a sin  ) ;
Now as A, B, C are the corresponding points on the auxiliary circle of

2

 a cos    3


2


 , a sin   
3



4


  and  a cos   
3



4


 , a sin   
3





Now equation of normal lines at P, Q, R are
ax sec   by cos ec  a 2  b2
2

ax sec   
3

2


  by cos ec   
3


 2
2
 a b

4

ax sec   
3

4


  by cos ec   
3



2
2
 a  b

To show that the lines are concurrent we observe the determinant D where D
2
a sec 
2

 a sec   
3

4

a sec   
3

b cos ec
2


 b cos ec   
3


4


 b cos ec   
3


a2  b2

2
2
 a b


2
2
 a b

sec 
cos ec
2
1
2 
2


  ab(a 2  b 2 ) sec   
 cos ec   
2
3 
3


4 
4


sec   
 cos ec   
3 
3


sin 
2

  sin   
3

4

sin   
3

cos 
2


 cos   
3


4


 cos   
3



 2


 2

sin 2
4 


 sin  2 

3 


8 


 sin  2 

3 


1
 ab(a 2  b 2 )
2
Where  
2 
2  
4


sin  cos  sin   
 cos   
 sin   
3 
3  
3


sin 
2

  sin   
3

2

sin   
3

cos 
2


 cos   
3


2


 cos   
3


4 


 cos   

3 


sin 2
4


 sin  2 
3


4


 sin  2 
3



4


 (**)  sin   
3






2


  sin   
3




 etc. 


(where  is the quantity outside the determinant )
 0 (apply R1  R1  R2  R3 )
5.
Let P and Q be (a cos  , b sin  ) and (a cos  , a sin  ) respectively. Then if R be ( h, k )
h
((a cos  ) s  (a cos  )r )
 a cos 
sr
, k
((b sin  ) s  (a sin  )r )
sr
We easily eliminate  from these equations to get the locus of R as
3
x 2 y 2 (r  s ) 2

1
a 2 (ar  bs)2
6.
Let P(a cos  , b sin  ) be any point on the ellipse
Then equation of line OP is y 
Equation of tangent at P is
Slope  
b sin 
x
a cos 
x2 y 2

 1.
a 2 b2
…(i)
x
y
cos   sin   1
a
b
b cos 
a sin 
 Equation of line perpendicular to this tangent and passing through the focus (ae, 0) is
y
a sin 
( x  ae)
b cos 
….(ii)
It is sufficient to show that abscissa of point of intersection of line in (i) and (ii) is
On equating (i) and (ii) we get
7.
a
e
b sin 
a sin 
x
( x  ae)
a cos 
b sin 

b2 x
 x  ae
a2

 b2 
a
x 1  2   ae  x.e2  ae  x 
e
 a 
x2 y 2
The ellipse is

1
25 4
Any tangent to this ellipse may be taken as y  mx  25m2  4
(*)
In order that (*) lies in the first quadrant we must take m  0 and  sign in the radical must be
chosen . Since (*) is a tangent to x 2  y 2  r 2 also , we must have c 2  r 2 (1  m2 )
 25m2  4  r 2 (1  m2 )
 m2 
r2  4
25  r 2
The given data is possible only when m2  0 i.e. 4  r 2  25 which is indeed given to us .
  
, 0  and (0,  )
 m 
Now , let (*) meet x-axis at A and y-axis at b then A and B are  
where   25m2  4.
4
Thus if ( h, k ) be the midpoint of AB then h  
Now 4k 2  25m2  4 

2m
,k 

2
on dividing we get m  
k
.
h
2
25k
 4.
h2
Thus Locus of ( h, k ) must be 4 x 2 y 2  4  25 y 2 .
8.
Tangent drawn at (t 2 , 2t ) of the parabola y 2  4ax
yy1  2a( x  x1 )
Now the ellipse is
or y.2t  2( x  t 2 )
(
a  1)
Or
2 x  2ty  2t 2  0
…(i)
x2 y 2

1
5
4
The equation of normal at ( 5 cos  , 2sin  ) is ax sec   by cos ec  a 2  b2
Or
5 sec   2 y cos ec  1
…..(ii)
Since (i) and (ii) are same lines we must have
Or
2 cos 
 t sin   2t 2
5
2
2t
2t 2


2 cos ec
1
5 sec 
…….…(*)
If t  0 then equating first and third ratio we get cos   0
    / 2 (there are exactly two points on the ellipse
x2 y 2

1
a 2 b2
whose eccentric angles  satisfy cos   0)
If t  0 then on equating second and third numbers in the double equality(*)
we get sin   2t . Also cos   t 2 5 on squaring and adding we get 
 (5t 2  1) (t 2  1)  0  t 2 
If t 
1
5
1
2
1
then sin   
, cos   
5
5
5
     tan 1 2 . If t  
4t 2  5t 4  1
 tan   2 (But  in the third quadrant )
1
1
1
then sin  
, cos   
5
5
5
 tan   2 with  in second quadrant      tan 1 2
9.
Let eccentric angle of the points where the line px  qy  r meets the ellipse be  and  then
x
  y  
 
cos
 sin
 cos
must be same as px  qy  r
a
2
b
2
2
5

  

   
 
 cos
 / a  sin
 / b cos
2 
2 

2


p
q
r
But as is given    

4
 cos
On substituting at (*)we get cos
 
 
2
2

 cos
(*)
 
2
 cos

4
ap cos  / 8
   bp cos  / 8
sin

r
2
r
On squaring and adding we get the required condition.
Objective Exercise 204-207
5.
x2 y 2

 1 then the equation
a 2 b2
ax sec  - by cosec  = a2 – b2 must be same as lx + my + n = 0
l
m
n
an
bn



 cos  
,sin  
2
2
a sec  b cos ec a 2  b2
l a b
m a 2  b2
If lx + my + n = 0 is a normal to the ellipse



12.
Square and add to get the condition given in choice (b)
Normal at P(a cos , b sin ) is ax sec  - by cosec  = a2 – b2
 a 2  b2 

a 2  b2 
 G is 
 and g is  0,
 etc.
 a sec  
 b cos ec 
Use (slope FB) (slope F′B) = -1 and b2 = a2(1 – e2)
14.
2ae  8,
8.
2a
 18 On multiplying we get 4a 2  18  8
e
Whence e 

 a 2  36  a  6
x2 y 2
2
 b 2  a 2 1  e 2   20  Equation of the ellipse is

1
3
36 20
IIT Page 207-209
1.
2.
3.
If r > 1 then 1 – r < 0, 1 + r > 0. The given equation implies (negative or zero) = 1 which is
impossible.
NOTE: If r < -1 then the equation is an ellipse and if -1 < r < 1 the equation represents a hyperbola
12 22
22 1
E 1, 2   
 1  0  1, 2  lies outside E, E  2,1 
  1  0  1, 2  lies inside E
9
4
9 4
C 1, 2   0  1, 2  lies inside C C  2,1  0   2,1 lies inside C.  only choice (d) fits.
For ellipse a2 = 16, b2 = 9, b2 = a2(1 – e2)  e 
7
4
 7,0 . Now radius of the circle
= distance between  7,0  and (0, 3) =  7  0    0  3
 Focus is (ae, 0) or
2
11.
2
4
By symmetry the quadrilateral is a rhombus .Therefore area is four times the area of the right
6
angled triangle formed by the tangent and axes in the 1st quadrant.
Now , ae  a2  b2  ae  2
12.
 5
 tangent ( in first quadrant ) at the end of latus rectum  2, 
 3
2 5 y
x
y
1 9
is
   1 i.e.
 1

Area = 4. . .3  27 sq .units .
9 3 5
9/ 2 3
2 2
x  3 3 cos  y  sin 
The equation of tangent at given  point is

1
27
1
x
y


 1  sum’s of the intercepts ‘s’ on axes is given by
3 3 sec cos ec
   ds
s  3 3 sec  cos ec    0,  ,
 3 3 sec tan   cos ec cot   0
 2  d
3 3 sin  cos
1
1



 tan 3  
 tan  
 
2
2
6
cos 
sin 
3 3
3
14.
Equation of auxiliary circle must be x 2  y 2  9 ……(1)

x y
 1
3 1
Equation of AM is
…….(2)
1
27
 12 9 
square units
,  Whence area of AOM  .OA  MN 
2
10
 5 5
From (1) and (2), we get M as  
15.
The ellips is

x2 y 2

1
16 4

equation of normal is 4 x sec θ  2 y cos ecθ  12
 7 cos θ

M is 
.sin θ    h, k 
 2

h
7 cos θ
2

4h 2
 k2 1
49
 cos θ 
2h
and k  sin θ
7
4 x2

locus of M is
…..(i).
 y2  1
49
3
4 3
e


Now e 2  1 
16 4
2

Equation of latus rectum of given ellipse is x  4 
solving (i) and (ii) we get
4
12  y 2  1
49

7
3
 2 3
2
y2  1
48 1


49 49
……(ii)
y
1
7
16.



required points are  2 3 
(D):
The ellipse is
1

7
 (C) is correct
x2 y 2

1
9
4
Equation of chord contact AB ( by
x  3y  3  0
……(i)
xx1 yy1

 1 ) is
a 2 b2
……..(ii)
Solve (i) & (ii) we get coordinates A and B as
 9 8
  , 
 5 5
17.
(C):
and (3, 0)
 (D) is correct.
Equation of altitude PE is y  4  3  x  3 and equation of altitude AD y 
Solving (i) & (ii) we get coordinates of orthocenter x 
18.
(A):
11
8
, y   (C) is correct
5
5
 h  3   k  4 
2
If the point is ( h, k ) we must have
8
5
2
 h  3k  3

2
10
10  h 2  6h  9  k 2  8k  16   h 2  9k 2  9  6hk  6h  18k
Required locus is 9 x 2  y 2  6 xy  54 x  62 y  241  0 ( a parabola )  (A) is correct.
20.
2 3 sin 3 θ
1
From the figure Area of PQR   2 3 sin θ  2sec θ  2cos θ  
cos θ
2
This decreases as θ increases since f '  θ   0
3
 15 
1
1
1
Now  h  1   cos θ 
 Max area 1 = 2 3  
  4
2
4
2
 4 

8
2 3  15. 15 45  5

1  45

16
8
5
3
 3
And Min. area  2 = 2 3  
  2
 2 

2 33 3 9
  8 2  36
4
2
8
Thus,
8
5
1  8 2  9
AIEEE/IIT MAINS Page 209-210
6.
(a)

Foci of the given ellipse are  7,0

Now radius of the circle = 4  equation of circle is  x2  y 2  6 y  7  0
7.
(c)
Given ellipse is
x2 y 2

1
6
2
… (1)
The equation of any tangent to it is y  2 x  6m2  2
… (2)
Also perpendicular to (2) through the center of ellipse is
y
1
x
m
… (3)
eliminating ‘m’ from (2) and (3) Gives the required locus as ( x2  y 2 )2  6x2  2 y 2
9