Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

If  P\left(\theta_1\right)  and  Q\left(\theta_2\right) are the extremities of any focal chord of the hyperbola  \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, then \cos ^2 \frac{\theta_1+\theta_2}{2}=\lambda \cos ^2 \frac{\theta_1-\theta_2}{2}, where \lambda  is equal to

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Equation of any chord joining the points  P\left(\theta_1\right)  and  Q\left(\theta_2\right) is,
\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 \left(\frac{\theta_1+\theta_2}{2}\right)If it passes through (ae, 0), then


\Rightarrow \quad \mathrm{e}^2 \cos ^2\left(\frac{\theta_1-\theta_2}{2}\right)=\cos ^2\left(\frac{\theta_1+\theta_2}{2}\right) \\
\Rightarrow \lambda=\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}

Posted by

Rakesh

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