L 1 : x c o s e c α − y s e c α = k c o t 2 α {L}_{1}: x cosec \alpha - ysec\alpha = kcot2\alpha L 1 : x cosec α − ysec α = k co t 2 α x s i n α − y c o s α = k cos 2 α sin 2 α \dfrac{x}{sin\alpha} - \dfrac{y}{cos\alpha} = k\dfrac{\cos 2\alpha}{\sin 2\alpha} s in α x − cos α y = k sin 2 α cos 2 α
x c o s α − y s i n α s i n α c o s α = k cos 2 α 2 s i n α c o s α \dfrac{xcos\alpha - ysin\alpha}{sin\alpha cos\alpha} = k\dfrac{\cos 2\alpha}{2sin\alpha cos\alpha} s in α cos α x cos α − ys in α = k 2 s in α cos α cos 2 α ⇒ x c o s α − y s i n α = k 2 cos 2 α ⇒ xcos\alpha - ysin\alpha = \dfrac{k}{2}\cos 2\alpha ⇒ x cos α − ys in α = 2 k cos 2 α
We know that Perpendicular distance from a point \left({x}_{1}, {y}_{1}\right)t o to t o i s ∣ a x 1 + b y 1 + c a 2 + b 2 ∣ is \left|\dfrac{a{x}_{1} + b{y}_{1} + c}{\sqrt{{a}^{2} + {b}^{2}}}\right| i s a 2 + b 2 a x 1 + b y 1 + c
Perpendicular distance from ( 0 , 0 ) ( 0, 0 ) ( 0 , 0 ) p = ∣ 0 − 0 − k 2 cos 2 α cos 2 α + sin 2 α ∣ p = \left|\dfrac{0 - 0 - \dfrac{k}{2}\cos 2\alpha}{\sqrt{\cos^{2}\alpha + \sin^{2}\alpha}}\right| p = cos 2 α + sin 2 α 0 − 0 − 2 k cos 2 α
⇒ p 2 = k 2 4 cos 2 2 α ⇒ {p}^{2} = \dfrac{{k}^{2}}{4}\cos^{2}2\alpha ⇒ p 2 = 4 k 2 cos 2 2 α
L 2 : x s i n α + y c o s α − k s i n 2 α = 0 {L}_{2}: xsin\alpha + ycos\alpha - ksin2\alpha = 0 L 2 : x s in α + ycos α − k s in 2 α = 0
Perpendicular distance from ( 0 , 0 ) ( 0, 0 ) ( 0 , 0 ) q = ∣ 0 + 0 − k s i n 2 α sin 2 α + cos 2 α ∣ q = \left|\dfrac{0 + 0 - ksin2\alpha}{\sqrt{\sin^{2}\alpha + \cos^{2}\alpha}}\right| q = sin 2 α + cos 2 α 0 + 0 − k s in 2 α
⇒ q 2 = k 2 sin 2 2 α ⇒ {q}^{2} = {k}^{2}\sin^{2}2\alpha ⇒ q 2 = k 2 sin 2 2 α
4 p 2 + q 2 = k 2 ( sin 2 2 α + cos 2 2 α ) 4{p}^{2} + {q}^{2} = {k}^{2}\left(\sin^{2}2\alpha + \cos^{2}2\alpha\right) 4 p 2 + q 2 = k 2 ( sin 2 2 α + cos 2 2 α )
We know that sin 2 t + cos 2 t = 1. \sin^{2}t + \cos^{2}t = 1 . sin 2 t + cos 2 t = 1. 4 p 2 + q 2 = k 2 4{p}^{2} + {q}^{2} = {k}^{2} 4 p 2 + q 2 = k 2
