The angle between the pair of lines x2+y2cos2⁡θ4+sin2⁡θ=x3−sin⁡θ2 is

sub θ=0 in given pair of lines then it becomes x2-3y2=0 where a=1,h=0,b=-3 Now cos⁡θ=a+b(a−b)2+2h2 =1-3(1+3)2+0 ⇒cos⁡θ=12⇒θ=π3

The angle between the pair of lines x2+y2cos2⁡θ4+sin2⁡θ=x3−sin⁡θ2 is

sub θ=0 in given pair of lines then it becomes x2-3y2=0 where a=1,h=0,b=-3 Now cos⁡θ=a+b(a−b)2+2h2 =1-3(1+3)2+0 ⇒cos⁡θ=12⇒θ=π3

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The angle between the pair of lines $(x^{2} + y^{2}) (\frac{\cos^{2} θ}{4} + \sin^{2} θ) = {(\frac{x}{\sqrt{3}} - \sin θ)}^{2}$ is

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$2 θ$

answer is C.

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Detailed Solution

$s u b θ = 0 i n g i v e n p a i r o f l i n e s t h e n i t b e c o m e s x^{2} - 3 y^{2} = 0 w h e r e a = 1, h = 0, b = - 3 N o w \cos θ = \frac{|a + b|}{\sqrt{(a - b)^{2} + (2 h^{2})}} = \frac{|1 - 3|}{\sqrt{{(1 + 3)}^{2} + 0}} \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$