Question

Show that the two straight lines `x^2(tan^2theta+cos^2theta)-2xytantheta+y^2sin^2theta=0` makes angles with the x-axis such that the difference of their tangents is 2.?

Answer 1

The given equation of a pair of straight lines is
`x^2(tan^2theta+cos^2theta)-2xytantheta+y^2sin^2theta=0`
`or,y^2-(2tantheta)/sin^2thetaxy+(tan^2theta+cos^2theta)/sin^2thetax^2=0`
Let the equations of component straight lines are `y-m_1x=0andy-m_2x=0`, where `m_1andm_2` are the slopes of the two lines.

So we can write

`(y-m_1x)(y-m_2x)=y^2-(2tantheta)/sin^2thetaxy+(tan^2theta+cos^2theta)/sin^2thetax^2`
`=>y^2-(m_1+m_2)xy+m_1m_2x^2=y^2-(2tantheta)/sin^2thetaxy+(tan^2theta+cos^2theta)/sin^2thetax^2`

Comparing both sides we get

`m_1+m_2=(2tantheta)/sin^2theta=(2sectheta)/ sintheta`

And

`m_1m_2=(tan^2theta+cos^2theta)/sin^2theta`

So

`(m_2-m_2)^2=(m_1+m_2)^2-4m_1m_2`

`=((2sectheta)/ sintheta)^2-(4(tan^2theta+cos^2theta))/sin^2theta`

`=4[(sec^2theta-tan^2theta-cos^2theta)/sin^2theta]`

`=4[(1-cos^2theta)/sin^2theta]`

`=4[(sin^2theta)/sin^2theta]`

`=4`

Hence `abs(m_1-m_2)=2`
So it proves that the difference of tangents of the angles made by the lines with the X-axis is 2