The equation 2x^2 + 4xy - ky^2 + 4x + 2y - 1 | vedclass

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(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with the given equation $2x^2 + 4xy - ky^

Vedclass · Accepted Answer

$-\frac{5}{3}$

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(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with the given equation $2x^2 + 4xy - ky^2 + 4x + 2y - 1 = 0$,we get $a = 2, h = 2, b = -k, g = 2, f = 1, c = -1$.For the equation to represent a pair of lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Substituting the values: $(2)(-k)(-1) + 2(1)(2)(2) - 2(1)^2 - (-k)(2)^2 - (-1)(2)^2 = 0$.$2k + 8 - 2 + 4k + 4 = 0$.$6k + 10 = 0$.$6k = -10$.$k = -\frac{10}{6} = -\frac{5}{3}$.

The equation $2x^2 + 4xy - ky^2 + 4x + 2y - 1 = 0$ represents a pair of lines. The value of $k$ is

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