JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Analyzing +4 | -1

\sin ^{-1} \theta

is the acute angle between the curves

x^2+y^2=4 x

and

x^2+y^2=8

(2,2)

, then

\theta=

View answer and explanation

Correct option: C

1 / \sqrt{2}

Option C (

1 / \sqrt{2}

) is correct because the angle between the curves at the intersection point is

\pi/4

, and

\sin^{-1}(1/\sqrt{2}) = \pi/4

More Questions from TANGENTS & NORMALS

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z