If β is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cos θ , 3 sin θ ) and (-3 sin θ , 3 cos θ) ; θ ϵ ( 0, (π/2) ); then (2 cot β/ sin 2 θ) is equal to

Question

If β is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cos θ , 3 sin θ ) and (-3 sin θ , 3 cos θ) ; θ ϵ ( 0, (π/2) ); then (2 cot β/ sin 2 θ) is equal to (A) (2/√3) (B) (1/√3) (C) √2 (D) (√3/4)

Tardigrade Admin · Accepted Answer

2 x d x+6 y d y=0 2 x d x=-c y d y -( d x / dy )=(3 y / x ) =stepe of normal m 1=(3 √3 sin θ/3 cos θ)=√3 tan θ m2=(3 √3 cos θ/-3 sin θ)-√3 cot θ tan β=[(√3 tan θ+√3 cot θ/1+(√3 tan θ)(-√3 cot θ))] tan β=√3( mid ( tan θ+ cot θ/1-3)) =(√3/2)( tan θ+ cot θ) text We have to find =(2 cot β/ sin 2 θ) -(22/√3( tan θ+ cot θ) 2 sin θ cos θ) =(2/√3(( sin θ) cos θ+( cos θ) sin θ) sin θ cos θ =(2/√3)

Q. If $β$ is one of the angles between the normals to the ellipse, $x^{2} + 3 y^{2} = 9$ at the points $(3 cos θ, 3 sin θ)$ and $(- 3 sin θ, 3 cos θ); θ ϵ (0, \frac{π}{2})$ ; then $\frac{2 c o t β}{s i n 2 θ}$ is equal to

$\frac{2}{3}$

$\frac{1}{3}$

$2$

$\frac{3}{4}$

Q. If β is one of the angles between the normals to the ellipse, x2+3y2=9 at the points (3cosθ,3sinθ) and (−3sinθ,3cosθ);θϵ(0,2π​); then sin2θ2cotβ​ is equal to

3​2​

3​1​

2​

43​​

Q. If $β$ is one of the angles between the normals to the ellipse, $x^{2} + 3 y^{2} = 9$ at the points $(3 cos θ, 3 sin θ)$ and $(- 3 sin θ, 3 cos θ); θ ϵ (0, \frac{π}{2})$ ; then $\frac{2 c o t β}{s i n 2 θ}$ is equal to

$\frac{2}{3}$

$\frac{1}{3}$

$2$

$\frac{3}{4}$