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If the pair of lines ax^2+2hxy+by^2= 0 (...

If the pair of lines `ax^2+2hxy+by^2= 0 (h^2 > ab)` forms an equilateral triangle with the line `lx + my + n=0` then `(a+3b)(3a+b)=`

A

`H^2`

B

`-H^2`

C

`2H^2`

D

`4H^2`

Text Solution

Verified by Experts

The correct Answer is:
D
Given `Ax^2+2Hxy+By^2=0` ....(i)
and `ax+by+c=0`.........(ii)
Since, triangle is equilateral , then angle between the two lines is `60^(@)`.
Now , angle between pair of lines is given by
`cos 60^(@)=(A+B)/(sqrt((A-B)^2+4H^2))`
`rArr (A+B)/(sqrt((A-B)^2+4H^2))=(1)/(2)`
`rArr (A-B)^2+4H^2=4(A+B)^2`
`rArr 4(A^2+B^2+2AB)-(A^2+B^2-2AB)=4H^2`
`rArr 3A^2+10AB+3B^2=4H^2`
`therefore (3A+B)(A+3B)=4H^2`.
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