Question

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

Answer 1

Hint: 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.
\[\begin{align}
  & {{x}^{2}}-3{{y}^{2}}-2x=8 \\
 & {{x}^{2}}-2x+1-3{{y}^{2}}-1=8 \\
 & \left( {{x}^{2}}-2x+1 \right)-3{{y}^{2}}=8+1 \\
 & {{\left( x-1 \right)}^{2}}-3{{y}^{2}}=9 \\
 & \dfrac{{{\left( x-1 \right)}^{2}}}{9}-\dfrac{3{{y}^{2}}}{9}=\dfrac{9}{9} \\
 & \dfrac{{{\left( x-1 \right)}^{2}}}{9}-\dfrac{{{y}^{2}}}{3}=1 \\
\end{align}\]
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
\[\begin{align}
  & e=\dfrac{\sqrt{9+3}}{3} \\
 & e=\dfrac{\sqrt{12}}{3} \\
 & e=\dfrac{2\sqrt{3}}{3} \\
 & e=\dfrac{2}{\sqrt{3}} \\
\end{align}\]
Foci are
\[\left( 1\pm 2\sqrt{3},\text{ 0} \right)\]
And the directrices are
\[\begin{align}
  & x-1=\pm \dfrac{3}{\dfrac{2}{\sqrt{3}}} \\
 & x-1=\pm \dfrac{3\sqrt{3}}{2} \\
 & x=1\pm \dfrac{3\sqrt{3}}{2} \\
\end{align}\]

Note: To calculate the various parameters of the hyperbola always check if the centre of the hyperbola is changed from the origin.