Question

Find the centre, eccentricity, foci and directrices of the hyperbola
(i) 16x² − 9y² + 32x + 36y − 164 = 0
(ii) x² − y² + 4x = 0
(iii) x² − 3y² − 2x = 8.

Answer 1

(i) The equation $16x^2-9y^2+32x+36y-164=0$ can be simplified in the following way:
$$\begin{gathered} 16(x^2+2x)-9\left(y^2-4y\right)=164 \\ \Rightarrow 16(x^2+2x+1)-9\left(y^2-4y+4\right)=164+16-36 \\ \Rightarrow 16(x+1{)}^2-9{\left(y-2\right)}^2=144 \\ \Rightarrow \frac{(x+1{)}^2}{9}-\frac{(y-2{)}^2}{16}=1 \end{gathered}$$
Thus, the centre is $\left(-1,2\right)$.
Eccentricity of the hyperbola = $\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{9+16}}{3}=\frac{5}{3}$
Foci = $\left(-1\pm 5,2\right)=\left(-6,2\right),\left(4,2\right)$

Equation of the directrices:
$$\begin{gathered} x+1=\pm \frac{a}{e} \\ \Rightarrow x=\pm \frac{3\times 3}{5}-1 \\ \Rightarrow x=\pm \frac{9}{5}-1 \\ \Rightarrow 5x-4=0\mathrm{or}5x+14=0 \end{gathered}$$

(ii) The equation $x^2-y^2+4x=0$ can be simplified in the following manner:
$$\begin{gathered} (x^2+4x+4)-y^2=4 \\ \Rightarrow {\left(x+2\right)}^2-y^2=4 \\ \Rightarrow \frac{(x+2{)}^2}{4}-\frac{y^2}{1}=1 \end{gathered}$$
Thus, the centre is $\left(-2,0\right)$.
Eccentricity of the hyperbola = $\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{4+1}}{2}=\frac{\sqrt{5}}{2}$
Foci = $\left(-2\pm \sqrt{5},0\right)$
Equation of the directrices:
$$\begin{gathered} x+2=\pm \frac{a}{e} \\ \Rightarrow x+2=\pm \sqrt{5} \end{gathered}$$

(iii) The equation $x^2-3y^2-2x=8$ can be simplified in the following manner:
$$\begin{gathered} (x^2-2x+1)-3y^2=8+1 \\ \Rightarrow {\left(x-1\right)}^2-3y^2=9 \\ \Rightarrow \frac{(x-1{)}^2}{9}-\frac{y^2}{3}=1 \end{gathered}$$
Thus, the centre is $\left(1,0\right)$.
Eccentricity of the hyperbola = $\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{9+3}}{3}=\frac{2\sqrt{3}}{3}$
Foci = $\left(1\pm 2\sqrt{3},0\right)$
Equation of the directrices:
$$\begin{gathered} x-1=\pm \frac{a}{e} \\ \Rightarrow x=1\pm \frac{2\sqrt{3}}{3} \end{gathered}$$