Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The equation of common tangents of the curves   and &nb

The equation of common tangents of the curves \mathrm{x^2+4 y^2=8}  and  \mathrm{y^2=4 x}  are

Option: 1

\mathrm{y-2 x-4=0, y+2 x+4=0}


Option: 2

\mathrm{y-2 x-2=0, y+2 x+2=0}


Option: 3

\mathrm{\quad 2 y-x-4=0,2 y+x+4=0}


Option: 4

\mathrm{2 y-x-2=0,2 y+x+2=0}


Answers (1)

best_answer

 Tangent in terms of  m ' for \mathrm{ y^2=4 x~ is~ y=m x+\frac{1}{m} \text { and for } \frac{x^2}{8}+\frac{y^2}{2}=1 \text { is } y=m x \pm}

\mathrm{\sqrt{8 m^2+2}}  For common tangents we must have \mathrm{\frac{1}{m}= \pm \sqrt{8 m^2+2} \Rightarrow m= \pm \frac{1}{2}}  thus common tangents are \mathrm{\mathrm{y}=\frac{\mathrm{x}}{2}+2 ~and ~\mathrm{y}=-\frac{\mathrm{x}}{2}-2} i.e. \mathrm{2 y-x-4=0}  and  \mathrm{2 y+x+4=0}

Posted by

Suraj Bhandari

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 Registration LIVE: NTA session 1 application soon at jeemain.nta.nic.in; BTech exam dates
anam-khan

JEE Main 2026 Registration LIVE: NTA session 1 applications open soon at jeemain.nta.nic.in; exam dates out
anam-khan

JEE Main 2026: Top NITs still competitive for PwD candidates, show BTech CSE closing ranks

Latest Question