Sum

Find k, the slope of one of the lines given by kx^{2} + 4xy - y^{2} = 0 exceeds the slope of the other by 8.

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#### Solution

Comparing the equation kx^{2} + 4xy - y^{2} = 0 with ax^{2} + 2hxy - by^{2} = 0, we get, a = k, 2h = 4, b = -1.

Let m_{1} and m_{2} be the slopes of the lines represented by kx^{2} + 4xy - y^{2} = 0

∴ m_{1} + m_{2} = `(-2"h")/"b" = - 4/(-1) = 4`

and m_{1}m_{2} = `"a"/"b" = "k"/(-1) = -"k"`

We are given that m_{2} = m_{1} + 8

∴ m_{1} + m_{1} + 8 = 4_{ }

∴ 2m_{1} = - 4 ∴ m_{1} = - 2 ...(1)

Also, m_{1}(m_{1} + 8) = - k

(-2)(- 2 + 8) = - k ...[By (1)]

∴ (-2)(6) = - k

∴ - 12 = - k

∴ k = 12

Concept: Homogeneous Equation of Degree Two

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