If `x^(2)+2hxy+y^(2)=0(hgt1)` represents the equation of the straight lines through the origin which make an angle `alpha` with the straight line `y+x=0`, then

Options

A. `sec 2 alpha = h`
B. `cos alpha = sqrt((1+h)/(2h))`
C. `m_(1)+m_(2) =- 2 sec 2 alpha`
D. `cot alpha = sqrt((h+1)/(h-1))`

Answer

A, B, C, D

Explanation

Let lines `x^(2) +2hxy +y^(2) =0` be given by `y = m_(1)x` and `y = m_(2)x`
`:. m_(1) +m_(2) =- 2h`
Slope of `y +x =0` is `-1`
`:. tan alpha = |(m_(1)+1)/(1-m_(1))|` and `tan alpha = |(m_(2)+1)/(1-m_(2))|`
`:. tan alpha =(m_(1)+1)/(1-m_(1))` and `tan alpha =- ((m_(2)+1)/(1-m_(2)))`
(for `+ve` signs, in both given the same value but `m_(1) ne m_(2))`.
`rArr m_(1) = (tan alpha -1)/(tan alpha +1), m_(2) = (tan alpha +1)/(tan alpha -1)`
`rArr m_(1) +m_(2) =- 2 sec 2 alpha`
`rArr h = sec 2 alpha`
`rArr cos 2 alpha = (1)/(h)`
`rArr2 cos^(2) alpha -1 =(1)/(h)`
`rArr cos alpha =((1+h)/(2h))^((1)/(2)) rArr cot alpha = sqrt((h+1)/(h-1))`