Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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.
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !