Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

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

Minimum area of a triangle formed by any tangent to the ellipse   \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}

with the coordinate axis is 

Option: 1

\mathrm{\frac{a^2+b^2}{2}}


Option: 2

\mathrm{\frac{(a+b)^2}{2}}


Option: 3

ab


Option: 4

\mathrm{\frac{(a-b)^2}{2}}


Answers (1)

best_answer

Equation of tangent at \mathrm{(a \cos \theta,}

\mathrm{b \sin \theta)}  is                                                                                           

 

\mathrm{\begin{aligned} & \frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1 \\\\ & P \equiv\left(\frac{a}{\cos \theta}, 0\right) Q \equiv\left(0, \frac{b}{\sin \theta}\right) \end{aligned}}

Area of OPQ = \mathrm{\frac{1}{2} \cdot\left(\frac{a}{\cos \theta}\right) \cdot\left(\frac{b}{\sin \theta}\right) \mid=\frac{a b}{|\sin 2 \theta|}}

\mathrm{(\text { Area })_{\min }=a b}.

Posted by

mansi

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 2025: BTech in AI gains popularity; IIT, NIT cut-offs for artificial intelligence
anam-khan

JEE Main 2025 Live: NTA JEE session 1 begins tomorrow; admit card link, exam pattern, passing marks
anam-khan

JEE Main 2025 exam centre changed at UAE's Sharjah due to “unavoidable circumstances”

Latest Question