If the pairs of straight lines `ax^2+2hxy-ay^2=0 and bx^2+2gxy-by^2=0` be such that each bisects the angles between the other , then prove that hg+abd=0.

Options

A. `hg+ab=0`
B. `ah+bg=0`
C. `h^(2)-ab=0`
D. `ag+bh=0`

Answer

A

Explanation

Since each pair of lines bisects the angles between the other. So, the equation of the bisectors of the angles bectween the lines given by `ax^(2)+2hxy-ay^(2)=0` is
`bx^(2)+2gxy-by^(2)=0`
The equation of the bisectors of the angles between the lines given by `ax^(2)+2hxy-ay^(2)=0` is
`(x^(2)-y^(2))/(a-(-a))=(xy)/(h)rArr hx^(2)-2axy-hy^(2)=0" (ii)"`
Since (i) and (ii) represent the same pair of straight lines.
`therefore" "(b)/(h)=(g)/(-a)=(-b)/(-h)rArr -ab=ghrArr ab+gh=0`