Mathematics Times – July 2019

common eccentricity e is given by

2

2 5 5

1

3 9 3

   

 

 

(5)

From (3) and (5), the foci of E are at the points

 

1

5

, 0

2

n

 

 

 

 

. each is at a distance
 

1

5

2

n

form the centre. For n9, this equals

5

16

which shows that (D) is false. For (C),

the length of the latus rectum ofEnis

2 1 2 4

3

n

n

b

b e  because of (5). by (4), this

equals

8 1

3 16 6





for n 9. Hence (c) is true.

Finally , the sides Rnare 2 an 1 and 2 bn 1. So its

area is 1 1

24

4

n n 2 n

a b   by(3) and (4). So the

series in (B) is the sum of a geometric

progression of N terms with the first term 12

and common ratio

1

2

. Its value is which is less

than

6.Sol:

1

1

24 1

2

 N

 

   

   

for every N. So (B) is

true.
6.Sol: (a,b,d)

We have, from the given information:

3 1

2 2

sin sin

R

P Q

   



3 1

sin , sin

2 2

P Q

i.e., P  60 or 120 ;  Q^30 or 150 

 P     60 ,Q 30 ,R^90 (Right triangle)

and P      60 ,Q 150 P Q 180 (not valid)

^ P      120 ,Q^30 R 30 (valid)

and P   120 ,Q 150 (Not valid)

Hence P      120 ,Q R 30 q r 1

now,

Slope of

1 / 4 1

3 3 / 4 3 3

RS

 

 

Equation:

3

3 3

2

x y

^

1 1

and

6 6

k OE

Length of

27 1 28 7

16 16 4 2

RS   

likewise

0 0 1

1 1 3

Area 0 1

2 6 48

3 1

1

4 4

SOE 



and radius of incircle of PQR is

1 1 3

(^3) 3(2 3)
2 2 2
5 3 2 3 2
1
2
r
   
   

