eccentricity of locus of hyperbola

Question

The locus of intersection of the lines $\sqrt 3 x-y-4\sqrt3 t=0$ and $\sqrt 3 tx+ty-4\sqrt 3=0$(where t is a parameter) is a hyperbola . we have to find its eccentricity .

I got the intersection point as $x=(2+2t^2)/(t)$ and $y=(2\sqrt 3)/(2\sqrt 3 t^2 +t)$.

after that I got stuck .

I'd say the locus of intersection of two lines in Euclidean space $;\Bbb R^n;$ is either the empty space or a point. What did you actually mean here? And the second equation doesn't seem to be a line. — DonAntonio, Apr 29 '17 at 08:53
It is a whole family of pairs of lines, so the locus of intersections can be pretty much anything. — Ivan Neretin, Apr 29 '17 at 09:16
@IvanNeretin yes , locus of intersection would be hyperbola . — a25bedc5-3d09-41b8-82fb-ea6c353d75ae, Apr 29 '17 at 09:30
@IvanNeretin Yes, after the edition of the question that $;t;$ is a parameter now that's clear... — DonAntonio, Apr 29 '17 at 09:58

score 1 · Accepted Answer · answered Apr 29 '17 at 10:27

The family of lines is

$$\begin{cases}&I\;\;\sqrt3\,x-y=4\sqrt3\,t\,/\,\cdot(-t)\implies-\sqrt3\,tx+ty=-4\sqrt3t^2\\{}\\ &II\;\;\sqrt3\,tx+ty=4\sqrt3\end{cases}\stackrel{\text{sum up}\,-tI+II}\implies$$

$$\;2ty=4\sqrt3(1-t^2)\implies \color{red}{y=\frac{2\sqrt3(1-t^2)}t}\stackrel{\text{substitute in I}}\implies$$

$$\sqrt3\,x=4\sqrt3\,t+\frac{2\sqrt3(1-t^2)}t\implies$$

$$ \color{red}{x=4t+\frac{2(1-t^2)}t=\frac{2(t^2+1)}t}$$

so

$$\begin{cases}&x^2=\cfrac{4(t^2+1)^2}{t^2}\\{}\\ &y^2=\cfrac{12(t^2-1)^2}{t^2}\end{cases}\implies 3x^2-y^2=48\implies\cfrac{x^2}{4^2}-\cfrac{y^2}{(4\sqrt3)^2}=1$$

and thus the eccentricity is

$$e=\frac{\sqrt{a^2+b^2}}a=\frac{\sqrt{16+48}}{4}=\frac{\sqrt{64}}{4}=2$$

eccentricity of locus of hyperbola

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