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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
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
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
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