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The locus of a point of intersection of two lines $x \sqrt{3}-y=k \sqrt{3}$ and
$\sqrt{3} k x+k y=\sqrt{3}, k \in R$, describes
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$\sqrt{3} k x+k y=\sqrt{3}, k \in R$, describes
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The correct answer is:
a hyperbola
We have $\begin{aligned} \quad x \sqrt{3}-y &=k \sqrt{3} ...(1)\\ \sqrt{3} k x+k y &=\sqrt{3} ...(2) \end{aligned}$
We will equate value of $k$ from eq. (1) and (2)
$\begin{aligned} & \frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3} x+y} \\ \therefore &(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3}) \\ \therefore & 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3} \text { i.e. } \frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1, \text { which is equation of hyperbola. } \end{aligned}$
We will equate value of $k$ from eq. (1) and (2)
$\begin{aligned} & \frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3} x+y} \\ \therefore &(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3}) \\ \therefore & 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3} \text { i.e. } \frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1, \text { which is equation of hyperbola. } \end{aligned}$
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