JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Analyzing +4 | -1

Let

\theta

be the acute angle between the tangents to the ellipse

\frac{{{x^2}}}{9} + \frac{{{y^2}}}{1} = 1

and the circle

{x^2} + {y^2} = 3

at their point of intersection in the first quadrant. Then,

\tan \,\,\theta

is equal to

View answer and explanation

Correct option: B

\frac{2}{{\sqrt 3 }}

Option B (

\frac{2}{{\sqrt 3 }}

) is correct because calculating the slopes of the tangents and using the angle between two lines formula gives

\tan \theta = \frac{2}{{\sqrt 3 }}

More Questions from TANGENTS & NORMALS

Practice All Chapter MCQs →

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