If ax^2-34xy-5y^2+2x+26y-5=0 represents a pai | vedclass

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

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$7$

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(A) The general equation of a second-degree curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if the determinant $\Delta = 0$,where $\Delta = \begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$. Comparing the given equation $ax^2-34xy-5y^2+2x+26y-5=0$ with the general form,we have $a=a$,$h=-17$,$b=-5$,$g=1$,$f=13$,and $c=-5$. Substituting these values into the determinant: $\begin{vmatrix} a & -17 & 1 \ -17 & -5 & 13 \ 1 & 13 & -5 \end{vmatrix} = 0$ Expanding along the first row: $a((-5)(-5) - (13)(13)) - (-17)((-17)(-5) - (13)(1)) + 1((-17)(13) - (-5)(1)) = 0$ $a(25 - 169) + 17(85 - 13) + 1(-221 + 5) = 0$ $a(-144) + 17(72) - 216 = 0$ $-144a + 1224 - 216 = 0$ $-144a + 1008 = 0$ $144a = 1008$ $a = \frac{1008}{144} = 7$ Thus,the value of $a$ is $7$.

If $ax^2-34xy-5y^2+2x+26y-5=0$ represents a pair of straight lines,then the value of $a$ is

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