Question

If the pair of straight lines x² – 2kxy – y² = 0 bisect the angle between the pair of straight lines x² – 2lxy – y² = 0, Show that the later pair also bisects the angle between the former

Answer 1

The equations of the given pair of straight lines are

x² – 2kxy – y² = 0  .......(1)

x² – 2lxy – y² = 0   .......(2)

Given that the pair x² – 2kxy – y² = 0 bisects the angle between the pair x² – 2lxy – y² = 0

∴ The equation of the bisector of the pair

x² – 2lxy – y² = 0 is the pair x² – 2kxy – y² = 0

The equation of the bisector of x² – 2lxy – y² = 0 is

`(x^2 - y^2)/(1 - ( - 1)) = (xy)/(- l)`

`(x^2 - y^2)/2 = (xy)/(- l)`

`x^2 - y^2 = - 2/l xy`

`x^2+2/l xy-y^2 =0`  .....(3)

Equation (3) and Equation (1) represents the same straight lines.

∴ The coefficients are proportional.

`1/1 = (2/l)/(- 2"k") = (- 1)/(- 1)`

1 = `1/((l"k")` = 1

`l"k"` = – 1  .....(4)

To show that the pair x² – 2lxy – y² = 0 bisects the angle between the pair x² – 2kxy – y² = 0, it is enough to prove the equation of the bisector of x² – 2kxy – y² = 0 is x² – 2lxy – y² = 0

The equation of the bisector of x² – 2kxy – y² = 0 is

`(x^2 - y^2)/(1 - ( - 1)) = (xy)/(- "k")`

`(x^2 - y^2)/2 = (xy)/(- "k")`

`x^2 - y^2 =  2/"k" xy`

`x^2 - y^2 = - 2/(1/l) xy`

By equationn (4)

x² – y² = 2lxy .

x² – 2lxy – y² = 0

∴ The pair x² – 2lxy – y² = 0 bisects the angle between the pair x² – 2kxy – y² = 0