Pair Of Straight Lines Question 118
Question: The equation of the pair of straight lines, each of which makes an angle $ \alpha $ with the line $ y=x $ , is
[MP PET 1990]
Options:
A) $ x^{2}+2xy\sec 2\alpha +y^{2}=0 $
B) $ x^{2}+2xycosec2\alpha +y^{2}=0 $
C) $ x^{2}-2xycosec2\alpha +y^{2}=0 $
D) $ x^{2}-2xy\sec 2\alpha +y^{2}=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
Any line through the origin is $ y=mx $ .
If it makes an angle $ \alpha $ with the line $ y=x $ , then we should have $ \tan \alpha =\pm { \frac{m_1-m_2}{1+m_1m_2} }=\pm \frac{(m-1)}{1+m} $ or $ {{(1+m)}^{2}}{{\tan }^{2}}\alpha ={{(m-1)}^{2}} $
$ \Rightarrow m^{2}-2m{ \frac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha } }+1=0 $
$ \Rightarrow m^{2}-2m\sec 2\alpha +1=0 $ , $ { \because \frac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }=\sec 2\alpha } $
But $ m=\frac{y}{x}, $ hence on eliminating m, we get the required equation $ y^{2}-2xy\sec 2\alpha +x^{2}=0 $
BETA
