Discussion Forum



1)

 If the equation ax2+2hxy+by2+2gx+2fy=0 has one line as the bisector of the angle  between co-ordinate axes, then 


A) $(a+b)^{2}=4(h^{2}+f^{2})$

B) $(a+b)^{2}=4(h^{2}+g^{2}+f^{2})$

C) $(a+b)^{2}=4h^{2}$

D) $(a+b)^{2}=4(h^{2}+g^{2})$

Answer:

Option C

Explanation:

Given , equation of line

ax2+2hxy+by2+2gx+2fy=0

 Let the slope of the line m1  and m2

 $\therefore$    $m_{1}+m_{2}=-\frac{2h}{b}$

$m_{1}m_{2}=\frac{a}{b}$

Given,  one line is bisector of angle between  coordinate axes.

  $\therefore$     $m_{1}=\tan\frac{\pi}{4}=1$

 $\Rightarrow$    1+m2=-2h/b   .........(i)

 $m_{2}=\frac{a}{b}$  ......(ii)

 From eqs. (i) and (ii) , we get

 $1+\frac{a}{b}=\frac{-2h}{b}$

 $\Rightarrow$   a+b=-2h

 (a+b)2  =4 h2