The equation of the perpendiculars drawn from | vedclass

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Question

(A) The general equation of a pair of straight lines is given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with the given equation $2x^2

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$6x^2 + 5xy + y^2 = 0$

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(A) The general equation of a pair of straight lines is given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with the given equation $2x^2 - 10xy + 12y^2 + 5x - 16y - 3 = 0$,we have $a = 2$,$2h = -10$ (so $h = -5$),and $b = 12$.The equation of the pair of lines passing through the origin and perpendicular to the lines represented by the homogeneous part of the given equation is $bx^2 - 2hxy + ay^2 = 0$.Substituting the values,we get $12x^2 - (-10)xy + 2y^2 = 0$,which simplifies to $12x^2 + 10xy + 2y^2 = 0$.Dividing the entire equation by $2$,we obtain $6x^2 + 5xy + y^2 = 0$.

The equation of the perpendiculars drawn from the origin to the lines represented by the equation $2x^2 - 10xy + 12y^2 + 5x - 16y - 3 = 0$ is

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