Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The minimum area of the triangle formed by any tangent to the ellipse<img alt="\mathrm{ \frac{x^2}{16}+\frac{y^2}{81}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cfrac

The minimum area of the triangle formed by any tangent to the ellipse\mathrm{ \frac{x^2}{16}+\frac{y^2}{81}=1} and the coordinate axes is

Option: 1

36


Option: 2

30


Option: 3

32


Option: 4

34


Answers (1)

best_answer

\mathrm{\text { An equation of tangent to the ellipse } \frac{x^2}{16}+\frac{y^2}{81}=1}

\mathrm{\frac{x}{4} \cos \theta+\frac{y}{9} \sin \theta=1}

\mathrm{\text { Coordinates of } A \text { and } B \text { are }(4 \sec \theta, 0) \text { and }(0,9 \operatorname{cosec} \theta)} respectively

\mathrm{\text { Area of } \triangle O A B=\frac{1}{2} \frac{36}{|\sin \theta \cos \theta|}=\frac{36}{\sin 2 \theta}}

So, minimum value of area is 36.

Posted by

Sanket Gandhi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2027 Registration Date LIVE: January session application form soon; city slip, admit card, exam dates
anam-khan

FIITJEE allegedly shut 32 centres, left 14,411 students without coaching: ED
anam-khan

JEE Main 2027 Application Form Highlights: Session 1 registration soon at jeemain.nta.nic.in

Latest Question