Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • C’ be a curve which is locus of point of intersection of lines x = 2 + m and my = 4 – m. A circle <img alt="s \equiv ( x-2 )^2 + ( y+ 1) ^2 = 25" src="https://entrancecorner.o

C’ be a curve which is locus of point of intersection of lines x = 2 + m and my = 4 – m. A circle

s \equiv ( x-2 )^2 + ( y+ 1) ^2 = 25intesects the curve C at four points P, Q, R and S. If O is

centre of the curve ‘C’, then OP^2 + OQ ^2 + OR ^2+ OS ^2 is

Option: 1

50


Option: 2

100


Option: 3

25


Option: 4

12


Answers (1)

best_answer

 

Rectangular Hyperbola -

x^{2}-y^{2}= a^{2}

- wherein

 

  

x -2 = m

               y + 1 = 4/m

          \therefore (x - 2) (y + 1) = 4

\Rightarrow XY = 4 , where \: X = x - 2, Y = y + 1

and      s \equiv ( x-2 )^2 + ( y+ 1) ^2 = 25         

\Rightarrow  X^2+ Y^2 = 25

            Curve ‘C’ & circle S both are concentric

            \         OP^2 + OQ ^2 + OR^2 + OS ^2 = 4r^2 = 4\cdot25 = 100

Posted by

Gunjita

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 2025: NIT Tiruchirappalli cut-offs rose for major BTech programmes this year; closing ranks
anam-khan

JoSAA 2025 reconstituted; NIT Rourkela to coordinate BTech counselling for IITs, NITs, IIITs
anam-khan

JEE Main 2025: BTech Information Technology cut-offs at NITs; know opening, closing ranks

Latest Question