If y=m₁x and y=m₂x be the 2 equations represented by ax²+2hxy+by²=0, then why is m₁.m₂=a/b

well the quadratic equation u posted represents a pair of lines.

So,

(y - m1x).(y - m2x) = 0  is the same as the posted quadratic...

y^2 -(m1 + m2)xy + m1m2x^2 = 0 compare this to the quadratic  ax²+2hxy+by²=0  (make coefficient of y 1 by dividing by b on both sides) =>  (a/b) x^2 + (2h/b)xy + y^2 = 0

now compare the coefficients of x

=> m1m2 = a/b

We have two st. line y - m1x = 0 and y - m2x =0. Therfore all the points lying on either of two st. lines will satisfy the equation (y - m1x)(y - m2x) = 0.

We acn rewrite the  equation as y² - (m1+m2)xy + m1m2x² =0 . .............. i)

eq. of pair of st. line is given by ax² + 2hxy +by² = 0 ..................  ii)

As equ i) and ii) represents the same pair of st. line so we must have,

1/b = m1m2/a = -(m1+m2)/2h

Hence m1 m2 = a/b

and m1 + m2 = - 2h/b