Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-14

Let xy = 4 and x^2 + y^2 = 4 have a common point in first quadrant, then at the common point

  • Option 1)

    Both intersect at \frac{\pi}{6}

  • Option 2)

    Both intersect at \frac{\pi}{}4

  • Option 3)

    Both intersect at \frac{\pi}{3}

  • Option 4)

    Both touch each other.

 

Answers (1)

best_answer

As we have learned

Condition for the two curves to touch -

Two curves touch each other if the tangents to each of them are parallel to each other.

so\:\;\theta =0

\therefore \:\;m_{1}=m_{2}

- wherein

Where m1  &  m2  are Tangents slopes at the point of intersection of two curves.

 

 On solving , we get (2,2) as common point in first quadrant 

Now xy=4 \Rightarrow x\frac{dy}{dx} + y=0 \Rightarrow \frac{dy}{dx}= -y/x

i.e \frac{dy}{dx} =-1  at (2,2) 

Also , x^{2}+y^{2}=4\Rightarrow 2x+2y\frac{dy}{dx} =0\Rightarrow \frac{dy}{dx}=-x/y

i.e  \frac{dy}{dx} =-1  at (2,2)

so at (2,2) both have same slope of tangents so both curves touch each other 

 

 

 

 

 

 


Option 1)

Both intersect at \frac{\pi}{6}

Option 2)

Both intersect at \frac{\pi}{}4

Option 3)

Both intersect at \frac{\pi}{3}

Option 4)

Both touch each other.

Posted by

Himanshu

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: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises
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

Latest Question