If the pair of lines x^2 - 2pxy - y^2=0 and x^2 - 2qxy - y^2=0 are such that each pair bisects the angle between the other pair, then pq equals.?
The solution is
The second pair must be identical with x^2 - y^2/1-(-1) =xy/-p
i.e with x^2 + (2/p)xy - y^2=0. Consequently, 2/p = -2q
I did not understand anything from these steps. PLease help
-
UP 0 DOWN 0 0 5
5 Answers
1Note : Given One pair of line bisect the angle between the other lines...
Means x^2 - 2pxy - y^2=0 is angle bisector of x^2 - 2qxy - y^2=0 and vice versa..
Now for ax^2 +2hxy + by^2 = 0 angle bisector is given as...
x^2 - y^2 / ( a-b) = xy / h
So for x^2 - 2pxy - y^2=0
ANgle bisector will be x^2 -y^2 / (2) = xy /(-p)
Means x^2 - y^2 = -2xy/p
or x^2 +2xy/p -y^2 = 0
Now compare it with
x^2 - 2qxy - y^2=0 .....because it also the angle bisector equation ....so they both must be identical
-2q = 2/p
or pq = -1
Ans
62@swordfish...
you can do taht by rotation of axis..
and the thing here is that the x and y used on the RHS and LHS are different..
ou should use x and X instead of x and X
62The funda is simple
x2-y2=(x-y)(x+y)
now the lines x-y and x+y are perpendicular .. so their direction can be chosen as X and Y
so we can rewrite the thing (ofcourse with some more work and a relation between h and a, b ) as x^2 - y^2 / ( a-b) = XY / h