If 2x^2 + 3xy - 2y^2 - 5x + fy - 3 = 0 repres | 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 $\De

Vedclass · Accepted Answer

$-\frac{25}{2}$

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 $2x^2 + 3xy - 2y^2 - 5x + fy - 3 = 0$ with the standard form:$a = 2, b = -2, c = -3$$2h = 3 \implies h = \frac{3}{2}$$2g = -5 \implies g = -\frac{5}{2}$$2f_{given} = f \implies f_{given} = \frac{f}{2}$Substituting these values into the condition $\Delta = 0$:$(2)(-2)(-3) + 2(\frac{f}{2})(-\frac{5}{2})(\frac{3}{2}) - 2(\frac{f}{2})^2 - (-2)(-\frac{5}{2})^2 - (-3)(\frac{3}{2})^2 = 0$$12 - \frac{15f}{4} - \frac{f^2}{2} + 2(\frac{25}{4}) + 3(\frac{9}{4}) = 0$$12 - \frac{15f}{4} - \frac{f^2}{2} + \frac{25}{2} + \frac{27}{4} = 0$Multiply by $4$ to clear denominators:$48 - 15f - 2f^2 + 50 + 27 = 0$$-2f^2 - 15f + 125 = 0$$2f^2 + 15f - 125 = 0$Factoring the quadratic equation:$2f^2 + 25f - 10f - 125 = 0$$f(2f + 25) - 5(2f + 25) = 0$$(f - 5)(2f + 25) = 0$Thus,$f = 5$ or $f = -\frac{25}{2}$.Comparing with the options,the correct value is $-\frac{25}{2}$.

If $2x^2 + 3xy - 2y^2 - 5x + fy - 3 = 0$ represents a pair of straight lines,then one of the possible values of $f$ is

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