Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the area of the region bounded by the ellipse x^2/16+y^2/9=1.

Q : 4 Find the area of the region bounded by the ellipse  \frac{x^2}{16}+\frac{y^2}{9}=1.

Answers (1)

best_answer

The area bounded by the ellipse : \frac{x^2}{16}+\frac{y^2}{9}=1.

Ellipse 5

Area will be 4 times the area of EAB.

Therefore,  Area\ of\ EAB= \int^4_{0} y dx

= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx   

= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx

= \frac{3}{4}\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]^4_{0}

= \frac{3}{4}\left [ 2\sqrt{16-16} +8\sin^{-1}(1)-0-8\sin^{-1}(0)\right ]

= \frac{3}{4}\left [ \frac{8\pi}{2} \right ]

= \frac{3}{4}\left [ 4\pi \right ] =3\pi

Therefore the area bounded by the ellipse will be = 4\times {3\pi} = 12\pi\ units.

 

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10, 12 board exams 2025 from April 7 to 11 for sports students
anam-khan

CBSE Class 12 history paper 2025 moderately difficult, tested students’ understanding
anam-khan

CBSE Class 12 Computer Science Exam Analysis 2025: Paper 'moderate', but 'tougher' than last three years

Latest Question