JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Analyzing +4 | -1

\beta

is one of the angles between the normals to the ellipse,

x^2+3 y^2=9

at the points

(3 \cos \theta, \sqrt{3} \sin \theta)

and

(-3 \sin \theta, \sqrt{3} \cos \theta) ; \in\left(0, \frac{\pi}{2}\right) ;

then

\frac{2 \cot \beta}{\sin 2 \theta}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct. By finding the slopes of the normals at the given points and using the tangent formula, the expression simplifies to

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

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