The number of integral values of p in the dom | vedclass

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(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. For this to represent a pair of lines,the determinant m

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$3$

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(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. For this to represent a pair of lines,the determinant must be zero: $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Comparing $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ with the general form,we have $a=2, h=2, b=-p, g=2, f=q/2, c=1$. Substituting these into the determinant condition: $2(-p)(1) + 2(q/2)(2)(2) - 2(q/2)^2 - (-p)(2)^2 - 1(2)^2 = 0$. $-2p + 4q - q^2/2 + 4p - 4 = 0$. Multiplying by $-2$: $q^2 - 8q - 4p + 8 = 0$. For $q$ to be an integer,the discriminant of this quadratic in $q$ must be a perfect square: $D = (-8)^2 - 4(1)(8 - 4p) = 64 - 32 + 16p = 32 + 16p = 16(2 + p)$. For $D$ to be a perfect square,$2 + p$ must be a perfect square $k^2$. Given $p \in [-5, 5]$,$2 + p$ can take values from $-3$ to $7$. The perfect squares in this range are $0, 1, 4$. $2 + p = 0 \Rightarrow p = -2$. $2 + p = 1 \Rightarrow p = -1$. $2 + p = 4 \Rightarrow p = 2$. Thus,the possible integral values of $p$ are $\{-2, -1, 2\}$. The number of such values is $3$.

The number of integral values of $p$ in the domain $[-5, 5]$ such that the equation $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ represents a pair of lines is:

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