Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The area of the rectangle formed by the perpendiculars from the center of the ellipse <img alt="\frac{x^{2}}{9}+\frac{y^{2}}{4}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5

The area of the rectangle formed by the perpendiculars from the center of the ellipse \frac{x^{2}}{9}+\frac{y^{2}}{4}=1to the tangent and normal at the point-whose eccentric angle is \pi / 4, is

Option: 1

\frac{30}{17}


Option: 2

\frac{30}{13}


Option: 3

\frac{27}{17}


Option: 4

\frac{27}{13}


Answers (1)

best_answer

 The given point \mathrm{P} is

\mathrm{\left(3 \cos \frac{\pi}{4}, 2 \sin \frac{\pi}{4}\right) \equiv\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)}.

The equation of tangent is \mathrm{\frac{x}{3 \sqrt{2}}+\frac{y}{2 \sqrt{2}}=1}  and that of normal is \mathrm{y-\sqrt{2}=\frac{3}{2}\left(x-\frac{3}{\sqrt{2}}\right)}

i.e., \mathrm{3 \sqrt{2} x-2 \sqrt{2} y-5=0}.

\mathrm{\mathrm{ON}=\frac{6 \sqrt{2}}{\sqrt{13}}} and \mathrm{\mathrm{OM}=\frac{5}{\sqrt{26}}}.
Thus area \mathrm{=\frac{30}{13}}. Hence \mathrm{(\mathrm{B})} is the correct answer.

Posted by

SANGALDEEP SINGH

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 Mains 2026 LIVE: NTA session 2 city slip soon; exam details, marking scheme, prep tips
anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare

Latest Question