IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

shilpanair

_{_}The angle between the straight lines x² - y²- 2y - 1 = 0

ruhi

hi shilpa....its a simple one...i ll tell u the formula
angle between the lines ax^2 +2hxy+by^2+2gx+2fy+c=0 is
angle = tan^ -1(2([h^2 - ab)^1/2]/(a+b))
hope u got it........ : )

vineetvsb

the angle is given by
tanx=[2(h^2-ab)^1/2]/(a+b)
thenx=tan(inv.)[2(h^2-ab)^1/2]/(a+b)
a=1,b=-1,h=0
x=tan(inv)[2(0-1)]/(0)
{ here [2(0-1)]/(0)=1 b'coz we hav to take modtherefore mod(-1) becomes=1}
x=tan(inv)[2/0]
x=tan(inv)

=x= (pi)/2
plz..... rate me

shilpanair

i did in the same way.the method is corrct.but the answer given is (pi)/2

abhijeet_0201

the ans will be pi/2 in the given lines
a=1,b=-1, h=0,
putting these values in formula,
angle(x)=tan(inv)[2(h^2-ab)^1/2]/(a+b)
x=tan(inv)[2(1-(-1))^1/2]/1-1
x=tan(Inv)(infinity)
therefore x=pi/2

i think u took b=1 instead of -1
if u find my sol correct plz plz rate me.

if wrong plz tell me

shilpanair

didn't noticed the minus sign.thanks a lot!

iitkgp_bipin

Compare it with ax²+2hxy+by²+2gx+2fy+c=0

Put the values in tan^-1[2(h²-ab)^1/2]/(a+b)

You will get

/2.

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