Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Tangents at right angles are drawn to ellipse <img alt="\mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E%7B2%7D%7D

Tangents at right angles are drawn to ellipse \mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}. The locus of mid points of chord of contact is \mathrm{\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)^{2}=k} then \mathrm{\mathrm{k}} is

Option: 1

1


Option: 2

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


Option: 3

\mathrm{\frac{x^2-y^2}{a^2+b^2}}


Option: 4

\mathrm{\frac{y^2-x^2}{a^2+b^2}}


Answers (1)

best_answer

Tangents at right angles to ellipse intersects on \mathrm{x^{2}+y^{2}=a^{2}+b^{2}}  any point on it \mathrm{\left(\sqrt{a^{2}+b^{2}} \cos \theta, \sqrt{a^{2}+b^{2}} \sin \theta\right)}  chord of contact \mathrm{\mathrm{P}}. with respect to ellipse \mathrm{\frac{x \cos \theta}{a^{2}}+\frac{y \sin \theta}{b^{2}}=\frac{1}{\sqrt{a^{2}+b^{2}}}}\quad \ldots \ldots(1)
Equation of chord having  \mathrm{\left(x_{1}, y_{1}\right)} as mid point is \mathrm{\frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}}\quad \ldots \ldots(2)

From (1) and (2) eliminate \mathrm{\theta} to get locus

Posted by

Deependra Verma

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 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question