Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The eccentric angles of the extremities of a chord of an 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%5

The eccentric angles of the extremities of a chord of an ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} are \mathrm{\theta_1} and \mathrm{\theta_2}. The chord cuts the major axis of the ellipse at a distance d from the centre. Prove that \mathrm{\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}=\mathrm{k}}, where \mathrm{\mathrm{k}=}

Option: 1

\mathrm{\frac{d-a}{d+a}}


Option: 2

\mathrm{\frac{d+a}{d-a}}


Option: 3

\mathrm{\frac{a-d}{a+d}}


Option: 4

\mathrm{\frac{a+d}{a-d}}


Answers (1)

best_answer

The equation of the chord joining \mathrm{\theta_1 \, \, and\, \, \theta_2} is

\mathrm{ \frac{\mathrm{x}}{\mathrm{a}} \cos \left(\frac{\theta_1+\theta_2}{2}\right)+\frac{\mathrm{y}}{\mathrm{b}} \sin \left(\frac{\theta_1+\theta_2}{2}\right)=\cos \frac{\theta_1-\theta_2}{2} }               ...(1)
It cuts the major axis at a distance d from the centre i.e. at \mathrm{(\mathrm{d}, 0)}

\mathrm{\therefore \frac{\mathrm{d}}{\mathrm{a}} \cos \frac{\theta_1+\theta_2}{2}=\cos \frac{\theta_1-\theta_2}{2} \quad \therefore \quad \frac{\mathrm{d}}{\mathrm{a}}=\frac{\cos \frac{\theta_1-\theta_2}{2}}{\cos \frac{\theta_1+\theta_2}{2}}}

Applying componendo and dividendo 

\mathrm{\frac{d-a}{d+a}=\frac{2 \sin \frac{\theta_1}{2} \sin \frac{\theta_2}{2}}{2 \cos \frac{\theta_1}{2} \cos \frac{\theta_2}{2}} \Rightarrow \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}=\frac{d-a}{d+a}}

 

Posted by

manish painkra

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question