MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
upendra hembram Grade:
        

find the greatest area of the rectangle which can be inscribed in the ellipse ; x2/a2+y2/b2=1 with sides parallel to the co-ordinate axes.

7 years ago

Answers : (1)

Askiitians Expert Bharath-IITD
23 Points
										

Dear Upendra,


Consider a rectangle as shown in the figure inscribed in the given ellipse


(x/a)2 + (y/b)2 = 1 such that its side are of length 2p and 2q


Now considerthe point (p,q) which is one of the ocrners of the rectangle that lies on the ellipse so


(p/a)2 + (q/b)2 = 1   which gives q= b * √(1-(p/a)2


now area of the rectangle A = 2p * 2q


                                        = 4 * p* q


                                        = 4 * p * b * √(1-(p/a)2


Now to find maximum area we find the maxima by differentiating it and equating it to zero


(dA/da) = 0


4*b*√(1-(p/a)2 + (4*p*b/√(1-(p/a)2) * (-p/a2) = 0


p = a/√2 and q = b/√2


Thus max area is found by substituting these vaues and comes out to be 2* a * b



Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.



All the best  !!!


 



Regards,


Askiitians Experts


Adapa Bharath 5628-2357_8384_pic.jpg

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 3,000 off
USE CODE: MOB20
Get extra R 320 off
USE CODE: MOB20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details