The equation of the line passing through pole and which is parallel to $r cos (θ - α) = p$ is

θ = \frac{π}{2} + α

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θ = \frac{π}{2} - α

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θ = \frac{π}{3} + α

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θ = \frac{π}{4} + α

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Solution

The correct option is C $θ = \frac{π}{2} + α$
Equation of line in polar form is given by
$cos (θ - α) = \frac{q}{r}$ , $q$ is the distance of the line from the pole
Since, it is paralllel to $r cos (θ - α) = p$
$⟹ r cos (θ - α) = 0$
$⟹ θ - α = \frac{π}{2}$
$⟹ θ = \frac{π}{2} + α$

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Similar questions

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

tan [\frac{π}{4} + θ] + tan [\frac{π}{4} - θ] = α

{tan}^{3} [\frac{π}{4} + θ] + {tan}^{3} [\frac{π}{4} - θ] =

sin (θ + α) = cos (θ + α)

θ, α \in (0, \frac{π}{2}) - {\frac{π}{4}}

P = [- sin (β - α), - cos β], Q = [cos (β - α), sin β]

R = [cos (β - α + θ), sin (β - θ)],

0 < α, β, θ < \frac{π}{4} .

P = (- sin (β - α), - cos β)

Q = (cos (β - α), sin β)

R = (cos (β - α + θ), sin (β - θ))

0 < α, β < \frac{π}{4}

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