The value of p for which the equation x^2+pxy | vedclass

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Question

(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 = abc+2f

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(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 $\Delta = abc+2fgh-af^2-bg^2-ch^2 = 0$. Comparing the given equation $x^2+pxy+y^2-5x-7y+6=0$ with the general form: $a=1, b=1, c=6, h=\frac{p}{2}, g=-\frac{5}{2}, f=-\frac{7}{2}$. Substituting these values into the condition $\Delta = 0$: $(1)(1)(6) + 2(-\frac{7}{2})(-\frac{5}{2})(\frac{p}{2}) - (1)(-\frac{7}{2})^2 - (1)(-\frac{5}{2})^2 - (6)(\frac{p}{2})^2 = 0$. $6 + \frac{35p}{4} - \frac{49}{4} - \frac{25}{4} - \frac{6p^2}{4} = 0$. Multiplying by $4$: $24 + 35p - 49 - 25 - 6p^2 = 0$. $-6p^2 + 35p - 50 = 0$. $6p^2 - 35p + 50 = 0$. $(2p-5)(3p-10) = 0$. Thus,$p = \frac{5}{2}$ or $p = \frac{10}{3}$.

The value of $p$ for which the equation $x^2+pxy+y^2-5x-7y+6=0$ represents a pair of straight lines is:

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