Notifications
Clear all
0
03/02/2022 6:53 pm
Topic starter
If p and q are the lengths of the perpendiculars from the origin on the lines,
x cosec α – y sec α = kcot 2α and
x sin α + y cosα = k sin 2α
respectively, then k2 is equal to :
(1) 4p2 + q2
(2) 2p2 + q2
(3) p2 + 2q2
(4) p2 + 4q2
1 Answer
0
03/02/2022 6:59 pm
Correct answer: (1) 4p2 + q2
Explanation:
First line is \(\frac{x}{sin α}\) - \(\frac{y}{cos α}\) = \(\frac{k\ cos 2α}{sin 2α}\)
⇒ xcosα - ysinα = \(\frac{k}{2}\)cos2α
⇒ p = \(|\frac{k}{2}cosα|\)
⇒ 2p = |kcos2α| .....(i)
second line is xsinα + ycosα = ksin2α
⇒ q = |ksin2α| .....(ii)
Hence 4p2 + q2 = k2 (From (i) & (ii))