Pair Of Straight Lines Question 50
Question: The equation of the locus of foot of perpendiculars drawn from the origin to the line passing through a fixed point (a, b), is
Options:
A) $ x^{2}+y^{2}-ax-by=0 $
B) $ x^{2}+y^{2}+ax+by=0 $
C) $ x^{2}+y^{2}-2ax-2by=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \lambda (x-a)+(y-b)=0 $ is the equation of line. $ r=-( \frac{-a\lambda -b}{{{\lambda }^{2}}+1} ) $
Coordinates of point $ \equiv { -\lambda ( \frac{-a\lambda -b}{{{\lambda }^{2}}+1} ),-( \frac{-a\lambda -b}{{{\lambda }^{2}}+1} ) } $
$ h=\lambda ( \frac{a\lambda +b}{{{\lambda }^{2}}+1} ),k=\frac{a\lambda +b}{{{\lambda }^{2}}+1},\lambda =\frac{h}{k} $
$ \therefore h=h( \frac{ah+kb}{h^{2}+k^{2}} )\Rightarrow x^{2}+y^{2}=ax+by. $
BETA
