Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If , then positive value of m for which <img alt="\mathrm{y=m x-b \sqrt{1+m^2}

If \mathrm{a>2 b>0}, then positive value of m for which \mathrm{y=m x-b \sqrt{1+m^2}} is a common tangent to \mathrm{x^2 +y^2=b^2} and  \mathrm{\left(x-(A)^2+y^2=b^2\right.}is
 

Option: 1

\mathrm{ \frac{2 b}{\sqrt{a^2-4 b^2}} }


Option: 2

\mathrm{ \frac{\sqrt{a^2-4 b^2}}{2 b} }


Option: 3

\mathrm{ \frac{2 b}{a-2 b} }


Option: 4

\mathrm{ \frac{b}{a-2 b} }


Answers (1)

best_answer

Given \mathrm{y=m x-b \sqrt{1+m^2}}  touches both the circles, so distance from the centre = radius of both the circles.

\begin{aligned} & \frac{\left|\mathrm{ma}-0-\mathrm{b} \sqrt{1+\mathrm{m}^2}\right|}{\sqrt{\mathrm{m}^2+1}}=\mathrm{b} \frac{\left|-\mathrm{b} \sqrt{1+\mathrm{m}^2}\right|}{\sqrt{\mathrm{m}^2+1}}=\mathrm{b} \Rightarrow\left|\mathrm{ma}-\mathrm{b} \sqrt{1+\mathrm{m}^2}\right|=\left|-\mathrm{b} \sqrt{1+\mathrm{m}^2}\right| \\ & \Rightarrow \mathrm{m}^2 \mathrm{a}^2-2 \mathrm{abm} \sqrt{1+\mathrm{m}^2}+\mathrm{b}^2\left(1+\mathrm{m}^2\right)=\mathrm{b}^2\left(1+\mathrm{m}^2\right) \\ & \Rightarrow \mathrm{ma}-2 \mathrm{~b} \sqrt{1+\mathrm{m}^2}=0 \Rightarrow \mathrm{m}^2 \mathrm{a}^2=4 \mathrm{~b}^2\left(1+\mathrm{m}^2\right) \Rightarrow \mathrm{m}=\frac{2 \mathrm{~b}}{\sqrt{\mathrm{a}^2-4 \mathrm{~b}^2}} \end{aligned}

 

Posted by

Gaurav

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 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question