The angle between y^2=4x and x^2+y^2=12 at a point of their intersection is

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Accepted Answer

Correct answer: C. (c) : To find out intersection point(s), ⇒x^2+y^2=12 ⇒x^2+4x=12 [Since y^2=4x is the 1^ st curve] ⇒x=2 ,-6 Clearly x=-6 does not exist for either first curve or second curve. At x=2 ,y=2 2 ⇒y^2=4x ⇒2y y_1=4 ⇒y_1=2/y ⇒y_1=1/ 2 [For 1^ st curve] Also, x^2+y^2=12 ⇒2x+2y y_1=0 ⇒y_1=-x/y =- 12 2 ⇒y_1=- 1 2 [For 2nd curve] ∴ Angle between curves = tan^-1 ⁡ ( m_1-m_21+m_1m_2 )= tan^-1 ⁡ ( 1 2- ( -1 2 )1+ ( 1 2 ) ( -1 2 ) ) = tan^-1 ⁡ ( 2 12 )= tan^-1 ⁡ (2 2 ).

$The angle between y^{2} = 4 x and x^{2} + y^{2} = 12 at a point of their intersection is$

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