If ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0 repr | vedclass

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(B) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $\Delta = 0$,where $\Delta =

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$4$

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(B) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $\Delta = 0$,where $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing the given equation $ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0$ with the general form,we have $a=a, h=3, b=b, g=-5, f=5, c=-6$.The condition $\Delta = 0$ gives:$\left|\begin{array}{ccc} a & 3 & -5 \ 3 & b & 5 \ -5 & 5 & -6 \end{array}ight| = 0$$a(-6b - 25) - 3(-18 + 25) - 5(15 + 5b) = 0$$-6ab - 25a - 21 - 75 - 25b = 0$$25a + 25b + 6ab + 96 = 0 \quad \dots(1)$For the lines to be perpendicular,the coefficient of $x^2$ plus the coefficient of $y^2$ must be zero,i.e.,$a + b = 0 \Rightarrow b = -a$.Substituting $b = -a$ into equation $(1)$:$25a + 25(-a) + 6a(-a) + 96 = 0$$-6a^2 + 96 = 0$$6a^2 = 96$$a^2 = 16$$|a| = 4$.

If $ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0$ represents a pair of perpendicular lines,then $|a| =$

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