An angle between the curves x^2-y^2=4 and x^2+y^2=4 2 is

Question

Accepted Answer

Correct answer: B. (b) Given that equation of curves are x^2-y^2=4 and x^2+y^2=4 2 Let (x_0 ,y_0 ) be point of intersection. Now, x^2-y^2=4 Differentiate w.r.t. ' x ' on both sides 2x-2y dy dx=0 ⇒ dy dx= xy At point (x_0 ,y_0 )m_1= x_0y_0 and curve x^2+y^2=4 2 Differentiate w.r.t. ' x ' on both sides 2x+2y dy dx=0 ⇒ dy dx=- xy At point (x_0 ,y_0 ) ,m_2= x_0y_0 Angle between tangents = tan^-1 ⁡ ( m_1-m_21+m_1m_2 ) ⇒ = tan^-1 ⁡ ( 2 x_0y_0 y_0^2-x_0^2y_0^2 ) ⇒ = tan^-1 ⁡ ( 2x_0y_04 ) [given x^2-y^2=4 satisfied point (x_0 ,y_0 ) ] = tan^-1 ⁡ ( x_0y_02 )……………..(i) Since, point (x_0 ,y_0 ) lie on both curves, so x_0^2-y_0^2=4 and x_0^2-y_0^2=4 2 ⇒x_0^2=2 (1+ 2 ) and y_0^2=2 ( 2-1 ) So, x_0^2=4 ( 2+1 ) ( 2-1 )=4 (2-1 )=4 x_0y_0=2……………(ii) Put in equation (i), = tan^-1 ( 22 ) = tan^-1 ⁡ (1 ) ⇒ = 4

$An angle between the curves x^{2} - y^{2} = 4 and x^{2} + y^{2} = 42 is$

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