The line `x-2y=0` will be a bisector of the angle between the lines represented by the equation `x^2-2hxy-2y^2=0` , if h is equal to

Options

A. `(1)/(2)`
B. `2`
C. `-2`
D. `-(1)/(2)` .

Answer

C

Explanation

Given , `x^2-2hxy-2y^2=0`......(i)
Equation of bsiector represnted by Eq.(i) is `(x^2-y^2)/(1+2)=(xy)/(-h)`
`rArr -hx^2-3xy+hy^2=0`
`rArr hx^2+3xy-hy^2=0` ....(ii)
Since , `x-2y=0` is one of the represented by Eq.(ii).
`therefore h(2y)^2+3(2y)y-hy^2=0`
`rArr 4h+6-h=0`
`rArr 3h+6=0`
`therefore h=-2` .