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Solve this problem - Limit , continuity and differentiability - JEE Main-23

The angle of intersection between x^2 + y^2 = 1 and x + y =1 is \theta at a point, where abscissa is positive, then \tan\theta equals ?

  • Option 1)

    1

  • Option 2)

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  • Option 3)

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  • Option 4)

    4

 
Answers (1)
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H Himanshu

As we have learned

Angle of intersection of two curves -

The angle of intersection of two curves is the angle subtended between the tangents at their point of intersection.Let  m1  &  m2 are two slope of tangents at intersection point of two curves then

tan\theta=\frac{[m_{1}-m_{2}]}{1+m_{1}m_{2}}

- wherein

where \theta is angle between two curves tangents.

 

 x^{2}+y^{2}=1   and  x+y =1 

\Rightarrow x^{2}+(1-x)^{2}=1 \Rightarrow 2x^{2}-2x=0 \Rightarrow 0=0,1

\therefore  point of intersection are (0,1) and (1,0)  we need to find angle at (1,0) 

x^{2}+y^{2}=1 \Rightarrow 2x +2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-x/y

\Rightarrow \frac{dy}{dx} at ( 1,0) will be infinite , i.e tangent will be vertical 

x+y=1\Rightarrow \frac{dy}{dx} =-1

\Rightarrow x+y =1   makes 135 \degree with x-axis so with a vertical line it will  make 45 \degree

\therefore \tan \theta = \tan 45 \degree =1 

 

 

 


Option 1)

1

Option 2)

2

Option 3)

3

Option 4)

4

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