Question

The locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}k=0$ and $\sqrt{3}kx+ky-4\sqrt{3}=0$ for different values of k is the hyperbola _______

Answer 1

Let point of intersection of lines is $(\alpha ,\beta )$

So, the point $(\alpha ,\beta )$ lies on both the lines.

$\therefore \sqrt{3}x-y-4\sqrt{3}k=0$

$\Rightarrow \sqrt{3}\alpha -\beta -4\sqrt{3}k=0$

$\Rightarrow k=\frac{\sqrt{3}\alpha -\beta }{4\sqrt{3}}\dots \left(1\right)$

And $\sqrt{3}kx+ky-4\sqrt{3}=0$

$\Rightarrow \sqrt{3}k\alpha +k\beta -4\sqrt{3}=0$

$\Rightarrow \sqrt{3}\alpha \times \frac{\sqrt{3}\alpha -\beta }{4\sqrt{3}}+\beta \times \frac{\sqrt{3}\alpha -\beta }{4\sqrt{3}}-4\sqrt{3}=0\left[\text{From equation}\right(1\left)\right]$

$\Rightarrow 3{\alpha }^2-\sqrt{3}\alpha \beta +\sqrt{3}\alpha \beta -{\beta }^2-48=0$

$\Rightarrow 3{\alpha }^2-{\beta }^2=48$

For the locus replace $\left(\alpha \beta \right)$ by $(x,y)$

$\Rightarrow 3x^2-y^2=48$

Hence, the locus of point of intersection of lines $\sqrt{3}x-y-4\sqrt{3}k=0$ and $\sqrt{3}kx+ky-4\sqrt{3}=0$ for the different value of $k$ is the hyperbola $3x^2-y^2=48$.