Question

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

Options

• `sqrt(2)`

• `2/sqrt(3)`

• `1/sqrt(3)`

• `sqrt(3)/4`

Answer 1

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

Explanation:

Since, x² + 3y² = 9

`\implies 2x + 6y (dy)/(dx)` = 0

`\implies (dy)/(dx) = (-x)/(3y)`

Slope of normal is `-(dx)/(dy) = (3y)/x`

`\implies (- (dx)/(dy))_((3 cos θ"," sqrt(3) sin θ)) = (3sqrt(3) sin θ)/(3 cos θ) = sqrt(3) tan θ` = m₁

and `(- (dx)/(dy))_((-3 sin θ"," sqrt(3) cos θ)) = (3sqrt(3) cos θ)/(-3 sin θ) = -sqrt(3) cot θ` = m₂

As, β is the angle between the normals to the given ellipse then

tan β = `|(m_1 - m_2)/(1 + m_1m_2)|`

= `|(sqrt(3) tan θ + sqrt(3) cot θ)/(1 - 3 tan θ cot θ)|`

= `|(sqrt(3) tan θ + sqrt(3) cot θ)/(1 - 3)|`

So, tan β = `sqrt(3)/2 |tan θ + cot θ|`

`\implies 1/cotβ = sqrt(3)/2 |sinθ/cosθ + cosθ/sinθ|`

`\implies 1/cotβ = sqrt(3)/2 |1/(sinθ cosθ)|`

`\implies 1/cotβ = sqrt(3)/(sin 2θ)`

`\implies (2 cot β)/(sin 2 θ)`

`\implies 2/sqrt(3)`