Question

Find the centre, eccentricity, foci and directrix of the hyperbola ${{x}^{2}}-3{{y}^{2}}-2x=8$.

Answer 1

To find the centre, eccentricity etc. convert the given form of the hyperbola into standard form.

Formula used:
For the hyperbola
$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$
The centre is $\left( h,\text{ }k \right)$.
Eccentricity is $\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}$
Foci is $\left( h\pm ae,\text{ }k \right)$ and
The directrice is $x-h=\pm \dfrac{ae}{e}$.

Complete step by step answer:
First transform the given equation into standard from.

& {{x}^{2}}-3{{y}^{2}}-2x=8 \\\ & {{x}^{2}}-2x+1-3{{y}^{2}}-1=8 \\\ & \left( {{x}^{2}}-2x+1 \right)-3{{y}^{2}}=8+1 \\\ & {{\left( x-1 \right)}^{2}}-3{{y}^{2}}=9 \\\ & \dfrac{{{\left( x-1 \right)}^{2}}}{9}-\dfrac{3{{y}^{2}}}{9}=\dfrac{9}{9} \\\ & \dfrac{{{\left( x-1 \right)}^{2}}}{9}-\dfrac{{{y}^{2}}}{3}=1 \\\ \end{aligned}$$ Here, $$h=1$$, $$k=0$$, $$a=3$$ and $$b=\sqrt{3}$$. Therefore, Centre of the hyperbola is $$\left( h,\text{ }k \right)=\left( 1,\text{ }0 \right)$$ The eccentricity is $$\begin{aligned} & e=\dfrac{\sqrt{9+3}}{3} \\\ & e=\dfrac{\sqrt{12}}{3} \\\ & e=\dfrac{2\sqrt{3}}{3} \\\ & e=\dfrac{2}{\sqrt{3}} \\\ \end{aligned}$$ Foci are $$\left( 1\pm 2\sqrt{3},\text{ 0} \right)$$ And the directrices are $$\begin{aligned} & x-1=\pm \dfrac{3}{\dfrac{2}{\sqrt{3}}} \\\ & x-1=\pm \dfrac{3\sqrt{3}}{2} \\\ & x=1\pm \dfrac{3\sqrt{3}}{2} \\\ \end{aligned}$$ **Note:** To calculate the various parameters of the hyperbola always check if the centre of the hyperbola is changed from the origin.