Question

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

• A. 7
• B. 5
• C. 2
• D. 13

Hint:

The general second-degree equation \(Ax^2+2Hxy+By^2+2Gx+2Fy+C=0\) represents a pair of straight lines if the determinant of the associated symmetric matrix is zero: \[ \begin{vmatrix} A & H & G
H & B & F
G & F & C \end{vmatrix} = 0 \]
Or, if \(ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0\).
Carefully identify A, B, C, F, G, H from the given equation.

Answer 1

The Correct Option is A

The general equation of a second degree \(S \equiv Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\) represents a pair of straight lines if the discriminant \(\Delta = 0\), where \[ \Delta = ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0 \] Alternatively, written as a determinant: \[ \Delta = \begin{vmatrix} A & H & G
H & B & F
G & F & C \end{vmatrix} = 0 \] Comparing the given equation \( ax^2 - 34xy - 5y^2 + 2x + 26y - 5 = 0 \) with the general form: \(A = a\) \(2H = -34 \implies H = -17\) \(B = -5\) \(2G = 2 \implies G = 1\) \(2F = 26 \implies F = 13\) \(C = -5\) Now substitute these into the determinant condition: \[ \begin{vmatrix} a & -17 & 1
-17 & -5 & 13
1 & 13 & -5 \end{vmatrix} = 0 \] Expand the determinant: \( a((-5)(-5) - (13)(13)) - (-17)((-17)(-5) - (13)(1)) + 1((-17)(13) - (-5)(1)) = 0 \) \( a(25 - 169) + 17(85 - 13) + 1(-221 - (-5)) = 0 \) \( a(-144) + 17(72) + (-221 + 5) = 0 \) \( -144a + 17 \times 72 - 216 = 0 \) \( 17 \times 72 = 17 \times (70+2) = 1190 + 34 = 1224 \). So, \( -144a + 1224 - 216 = 0 \) \( -144a + 1008 = 0 \) \( 144a = 1008 \) \( a = \frac{1008}{144} \) We can simplify this fraction. \(1008 / 12 = 84\). \(144 / 12 = 12\). So \(a = 84/12\). \(84 / 12 = 7\). So, \(a = 7\). This matches option (a). \[ \boxed{7} \]