Question

The locus of the point of intersection of the lines $$\begin{gathered} \sqrt{3x}-\mathrm{y}-4\sqrt{3t}=0 \\ \sqrt{3tx}+\mathrm{Ty}-4\sqrt{3}=0 \end{gathered}$$ (where T is a parameter) is a hyperbola. whose eccentricity is

Answer 1

Step 1: To derive the Standard equation of a hyperbola

Given:
$\sqrt{3}\mathrm{x}-\mathrm{y}-4\sqrt{3}\mathrm{t}=0$
$\sqrt{3}\mathrm{tx}+\mathrm{ty}-4\sqrt{3}=0$

The equations of lines
$\sqrt{3}\mathrm{x}-\mathrm{y}-4\sqrt{3}\mathrm{t}=0$
$\sqrt{3}\mathrm{x}-\mathrm{y}-4\sqrt{3}\mathrm{t}=0$
$\mathrm{t}=$$\frac{\sqrt{3\mathrm{x}-\mathrm{y}}}{4\sqrt{3}}-------\left[1\right]$
$\sqrt{3}\mathrm{tx}+\mathrm{ty}-4\sqrt{3}=0$
$\mathrm{t}(\sqrt{3}\mathrm{x}+\mathrm{y})=\frac{4\sqrt{3}}{\sqrt{3}\mathrm{x}+\mathrm{y}}----\left[2\right]$

Multiplying the equations $\left(1\right)\mathrm{and}\left(2\right)$

$\begin{array}{c}3{\mathrm{tx}}^2-{\mathrm{ty}}^2=48\mathrm{t}\\ \frac{3{\mathrm{tx}}^2}{48\mathrm{t}}-\frac{{\mathrm{ty}}^2}{48\mathrm{t}}=1\\ \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{48}=1\end{array}$

Standard equation of a hyperbola, $\Rightarrow \frac{{\mathrm{x}}^2}{a}-\frac{{\mathrm{y}}^2}{b}=1$ where ${\mathrm{a}}^2=16$ and ${\mathrm{b}}^2=48$

Step 2 : Compute the eccentricity

Eccentricity formula:

$\begin{array}{rl}& \mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}}\\ & \mathrm{e}=\sqrt{\frac{16+48}{16}}\\ & \mathrm{e}=\frac{8}{4}\\ & \mathrm{e}=2\end{array}$

Therefore, the eccentricity is $e=2$.