If the lines represented by the equation `2x^2-3xy+y^2=0` make angles `alpha` and `beta` with X-axis, then `cot^2alpha+cot^2beta` is equal to

Options

A. 0
B. `(3)/(2)`
C. `(7)/(4)`
D. `(5)/(4)`

Answer

D

Explanation

Given, `2x^2-3xy+y^2=0`.......(i)
Let the represented by Eq.(i) makes angle `alpha` and `beta` with X-axis
`therefore tanalpha=m_1`
and `tan beta=m_2`
From Eq.(i), we get
`m_1+m_2=tanalpha+tanbeta=(-2h)/(b)=(3)/(2)=3`
and `m_1.m_2=tanalpha.tanbeta=(ab)/(b)=2`
Now , `cot^2alpha+cot^2beta=(1)/(tan^2alpha)+(1)/(tan^2beta)=(tan^2alpha+tan^2beta)/(tan^2alpha.tan^2beta)`
`=((tanalpha+tanbeta)^2-2tanalpha tanbeta)/(tan^2alpha tan ^2beta)`
`=((3)^2-2xx2)/((2)^2)=(9-4)/(4)=(5)/(4)`