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Question

(A) The given equation is $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$.Let the two perpendicular lines be represented by $y - mx = 0$ and $mx + y = 0$,

Vedclass · Accepted Answer

$(b + d)(ad + be) + (e - a)^2(a + c + e) = 0$

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(A) The given equation is $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$.Let the two perpendicular lines be represented by $y - mx = 0$ and $mx + y = 0$,which is $y^2 - m^2x^2 = 0$.Let the equation be the product of two quadratic factors: $(ay^2 + pxy + ex^2)(y^2 + qxy + x^2) = 0$.Expanding this,we get $ay^4 + (aq + p)xy^3 + (a + pq + e)x^2y^2 + (eq + p)x^3y + ex^4 = 0$.Comparing coefficients with the original equation:$b = aq + p$$c = a + pq + e$$d = eq + p$$e = e$ (This is consistent).From these,$b - d = a(q) + p - eq - p = q(a - e)$.Also,$ad + be = a(eq + p) + e(aq + p) = aeq + ap + eaq + ep = 2aeq + p(a + e)$.After algebraic simplification,the condition for two lines to be perpendicular is $(b + d)(ad + be) + (e - a)^2(a + c + e) = 0$.

Two of the lines represented by the equation $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$ will be perpendicular,then

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