The locus of the point of intersection of the | vedclass

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(A) Given lines are:$L_1: \sqrt{2}x - y + 4\sqrt{2}k = 0 \Rightarrow y = \sqrt{2}x + 4\sqrt{2}k \quad (i)$$L_2: \sqrt{2}kx + ky - 4\sqrt{2} = 0 \quad

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$A$ hyperbola with length of its transverse axis $8\sqrt{2}$

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(A) Given lines are:$L_1: \sqrt{2}x - y + 4\sqrt{2}k = 0 \Rightarrow y = \sqrt{2}x + 4\sqrt{2}k \quad (i)$$L_2: \sqrt{2}kx + ky - 4\sqrt{2} = 0 \quad (ii)$Substituting $y$ from $(i)$ into $(ii)$:$\sqrt{2}kx + k(\sqrt{2}x + 4\sqrt{2}k) - 4\sqrt{2} = 0$$\sqrt{2}kx + \sqrt{2}kx + 4\sqrt{2}k^2 - 4\sqrt{2} = 0$$2\sqrt{2}kx = 4\sqrt{2}(1 - k^2)$$x = \frac{2(1 - k^2)}{k}$Substituting $x$ back into $(i)$:$y = \sqrt{2}(\frac{2(1 - k^2)}{k}) + 4\sqrt{2}k = \frac{2\sqrt{2} - 2\sqrt{2}k^2 + 4\sqrt{2}k^2}{k} = \frac{2\sqrt{2}(1 + k^2)}{k}$From these,we observe the relationship between $x$ and $y$:$\frac{y}{4\sqrt{2}} = \frac{1+k^2}{2k}$ and $\frac{x}{4} = \frac{1-k^2}{2k}$$(\frac{y}{4\sqrt{2}})^2 - (\frac{x}{4})^2 = (\frac{1+k^2}{2k})^2 - (\frac{1-k^2}{2k})^2 = \frac{(1+k^2)^2 - (1-k^2)^2}{4k^2} = \frac{4k^2}{4k^2} = 1$This is the equation of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ where $a = 4\sqrt{2}$.The length of the transverse axis is $2a = 2(4\sqrt{2}) = 8\sqrt{2}$.

The locus of the point of intersection of the lines,$\sqrt{2}x - y + 4\sqrt{2}k = 0$ and $\sqrt{2}kx + ky - 4\sqrt{2} = 0$ (where $k$ is any non-zero real parameter) is

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