If the equation lambda x^2 - 10xy + 12y^2 + 5 | vedclass

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Question

(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with the given equation $\lambda x^2 - 1

Vedclass · Accepted Answer

$2$

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(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with the given equation $\lambda x^2 - 10xy + 12y^2 + 5x - 16y - 3 = 0$,we get: $a = \lambda, h = -5, b = 12, g = 5/2, f = -8, c = -3$. For the equation to represent a pair of straight lines,the condition $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ must be satisfied. Substituting the values: $\lambda(12)(-3) + 2(-8)(5/2)(-5) - \lambda(-8)^2 - 12(5/2)^2 - (-3)(-5)^2 = 0$ $-36\lambda + 400 - 64\lambda - 75 + 75 = 0$ $-100\lambda + 400 = 0$ $100\lambda = 400$ $\lambda = 4$. Wait,re-evaluating the determinant condition $\Delta = \begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$: $\begin{vmatrix} \lambda & -5 & 5/2 \ -5 & 12 & -8 \ 5/2 & -8 & -3 \end{vmatrix} = 0$ $\lambda(12(-3) - (-8)(-8)) - (-5)((-5)(-3) - (-8)(5/2)) + (5/2)((-5)(-8) - 12(5/2)) = 0$ $\lambda(-36 - 64) + 5(15 + 20) + (5/2)(40 - 30) = 0$ $-100\lambda + 5(35) + (5/2)(10) = 0$ $-100\lambda + 175 + 25 = 0$ $-100\lambda + 200 = 0$ $\lambda = 2$.

If the equation $\lambda x^2 - 10xy + 12y^2 + 5x - 16y - 3 = 0$ represents a pair of straight lines,then what is the value of $\lambda$?

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