If the equation x^2 - ky^2 - 4x + 6y - 5 = 0 | vedclass

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(B) Let $f(x, y) = x^2 - ky^2 - 4x + 6y - 5 = 0$.For the equation to represent a pair of straight lines, the partial derivatives with respect to $x$ a

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$(2, 3)$

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(B) Let $f(x, y) = x^2 - ky^2 - 4x + 6y - 5 = 0$.For the equation to represent a pair of straight lines, the partial derivatives with respect to $x$ and $y$ must be zero at the point of intersection $(h, k_0)$.$\frac{\partial f}{\partial x} = 2x - 4 = 0 \implies x = 2$.$\frac{\partial f}{\partial y} = -2ky + 6 = 0 \implies ky = 3$.Also, the condition for a pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Here $a=1, b=-k, c=-5, h=0, g=-2, f=3$.Substituting these: $(1)(-k)(-5) + 2(3)(-2)(0) - 1(3)^2 - (-k)(-2)^2 - (-5)(0)^2 = 0$.$5k - 9 + 4k = 0 \implies 9k = 9 \implies k = 1$.Substituting $k=1$ into $ky=3$, we get $y=3$.Thus, the point of intersection is $(2, 3)$.

If the equation $x^2 - ky^2 - 4x + 6y - 5 = 0$ represents a pair of straight lines, then their point of intersection is

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