Question

The locus of the point of intersection of the lines $\sqrt{3}kx+ky-4\sqrt{3}=0$ and $\sqrt{3}x-y-4\sqrt{3}k=0$ is a conic, whose length of latus rectum is equal to

Answer 1

Given, $\sqrt{3}kx+ky=4\sqrt{3}\dots \left(1\right)$
$\sqrt{3}x-y=4\sqrt{3}k$
$\Rightarrow \sqrt{3}kx-ky=4\sqrt{3}{k}^2\dots \left(2\right)$
From $\left(1\right)+\left(2\right)$
$2\sqrt{3}kx=4\sqrt{3}(1+k^2)$
$\Rightarrow x=2(k+\frac{1}{k})\dots \left(3\right)$
From $\left(1\right)-\left(2\right)$
$2ky=4\sqrt{3}(1-k^2)$
$\Rightarrow y=2\sqrt{3}(\frac{1}{k}-k)\dots \left(4\right)$
From $\left(3\right)$ and $\left(4\right)$
$\therefore {\left(\frac{x}{2}\right)}^2-{\left(\frac{y}{2\sqrt{3}}\right)}^2={(\frac{1}{k}+k)}^2-{(\frac{1}{k}-k)}^2$
$\Rightarrow {\left(\frac{x}{2}\right)}^2-{\left(\frac{y}{2\sqrt{3}}\right)}^2=4$
$\Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1$ (Hyperbola)
$a^2=16$ and $b^2=48$
Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 48}{4}=24$