Prove that the product of the perpendiculars from `(alpha,beta)` to the pair of lines `a x^2+2h x y+b y^2=0` is `(aalpha^2-2halphabeta+bbeta^2)/(sqrt((a-b)^2+4h^2))`

Explanation

Let the lines be `y=m_(1)xandy=m_(2)x`
or `m_(1)x-y=0andm_(2)x-y=0`
were `m_(1)+m_(2)=-(2h)/(b),m_(1)m_(2)=(a)/(b)`
Now , if the lengths of perpendicular from (0,0) on these lines are `d_(1)andd_(2)` respectively , then
`d_(1)d_(2)=(|m_(1)alpha-beta|)/(sqrt(m_(1)^(2)+1))(|m_(2)alpha-beta|)/(sqrt(m_(2)^(2)+1))`
`=(|m_(1)m_(2)alpha^(2)-(m_(1)+m_(2))alpha beta+beta^(2)|)/(sqrt(m_(1)^(2)m_(2)^(2)+m_(1)^(2)+m_(2)^(2)+1))`
`=(|m_(1)m_(2)alpha^(2)-(m_(1)+m_(2))alpha beta+beta^(2)|)/(sqrt(m_(1)^(2)m_(2)^(2)+(m_(1)+m_(2))^(2)-2m_(1)m_(2)+1))`
`(|(a)/(b)alpha^(2)+(2h)/(b)alpha beta+ beta^(2)|)/(sqrt(a^(2)/(b^(2))+(4h^(2))/(b^(2))-2(a)/(b)+1))`
`=(|aalpha+2halpha beta+b beta^(2)|)/(sqrt(a^(2)+4h^(2)-2ab+b^(2)))`
`=(|aalpha+2halpha beta+b beta^(2)|)/(sqrt((a-b)^(2)+4h^(2)))`