Pair Of Straight Lines Question 137
Question: The lines represented by the equation $ ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 $ will be equidistant from the origin, if
Options:
A) $ f^{2}+g^{2}=c(b-a) $
B) $ f^{4}+g^{4}=c(bf^{2}+ag^{2}) $
C) $ f^{4}-g^{4}=c(bf^{2}-ag^{2}) $
D) $ f^{2}+g^{2}=af^{2}+bg^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let the equations represented by $ ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 $ be $ lx+my+n=0 $ and $ l’x+m’y+n’=0 $ .
Then the combined equation represented by these lines is given by $ (lx+my+n)(l’x+m’y+n’)=0 $
So, it must be similar with the given equation.
On comparing, we get $ ll’=a,\ mm’=b\ \ nn’=c,\ \ lm’+ml’=2h,\ \ ln’+l’n=2g $ , $ mn’+nm’=2f $
According to the condition, the length of perpendiculars drawn from origin to the lines are same.
So, $ \frac{n}{\sqrt{l^{2}+m^{2}}}=\frac{n’}{\sqrt{l{{’}^{2}}+m{{’}^{2}}}}=\frac{{{(nn’)}^{2}}}{(l^{2}+m^{2})(l{{’}^{2}}+m{{’}^{2}})} $ Now on eliminating $ l,\ m,\ l’,\ m’ $ and $ n,\ n’ $ ,
we get the required condition $ f^{4}-g^{4}=c(bf^{2}-ag^{2}). $
BETA
