Question: If p and $ p’ $ be the distances of origin from the lines $ x\sec \alpha +y\text{cosec }\alpha =k $ and $ x\cos \alpha -y\sin \alpha =k\cos 2\alpha $ , then $ 4p^{2}+{{{p}’}^{2}} $ =
Options:
A)k
B) $ 2k $
C) $ k^{2} $
D) $ 2k^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Here $ p=| \frac{-k}{\sqrt{{{\sec }^{2}}\alpha +cose{c^{2}}\alpha }} | $ , $ {p}’=| \frac{-k\cos 2\alpha }{\sqrt{{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha }} | $ Hence $ 4p^{2}+{{{p}’}^{2}}=\frac{4k^{2}}{{{\sec }^{2}}\alpha +cose{c^{2}}\alpha }+\frac{k^{2}{{({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )}^{2}}}{1} $ $ =4k^{2}{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha +k^{2}({{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha ) $ $ -2k^{2}{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha $ $ =k^{2}{{({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )}^{2}}=k^{2} $ .
BETA
