Find the eccentricity, center, vertices, foci, length of latusrectum and the equations of directrices of the hyperbolas
$$\begin{gathered} \text{ i) }5{\mathrm{x}}^2-4{\mathrm{y}}^2+20\mathrm{x}+8\mathrm{y}-4=0 \\ \text{ ii) }4{\mathrm{x}}^2-9{\mathrm{y}}^2-8\mathrm{x}-32=0 \end{gathered}$$
iii)Find the eccentricity, length of latus rectum, centre, foci, vertices and the equation to the directrices of the hyperbola
iv) Find the eccentricity, length of latus rectum, centre, foci, vertices and the equation to the directrices of the hyperbola
$\begin{array}{l}5{\mathrm{x}}^2-4{\mathrm{y}}^2+20\mathrm{x}+8\mathrm{y}-4=0\\ 4{\mathrm{x}}^2-5{\mathrm{y}}^2-16\mathrm{x}+10\mathrm{y}+31=0\end{array}$

$\begin{array}{l}\text{ i) }5{\mathrm{x}}^2-4{\mathrm{y}}^2+20\mathrm{x}+8\mathrm{y}-4=0\\ 5{\mathrm{x}}^2+20\mathrm{x}-4{\mathrm{y}}^2+8\mathrm{y}-4=0\\ 5\left({\mathrm{x}}^2+4\mathrm{x}\right)-4\left({\mathrm{y}}^2-2\mathrm{y}+1\right)=0\\ 5\left({\mathrm{x}}^2+2\left(2\right)\mathrm{x}+4-4\right)-4(\mathrm{y}-1{)}^2=0\\ 5(\mathrm{x}+2{)}^2-20-4(\mathrm{y}-1{)}^2=0\\ 5(\mathrm{x}+2{)}^2-4(\mathrm{y}-1{)}^2=20\\ \frac{5}{20}(\mathrm{x}+2{)}^2-\frac{4}{20}(\mathrm{y}-1{)}^2=1\\ \frac{(\mathrm{x}+2{)}^2}{4}-\frac{(\mathrm{y}-1{)}^2}{5}=1\end{array}$
This is in the form of $\frac{(\mathrm{x}-\mathrm{h}{)}^2}{{\mathrm{a}}^2}-\frac{(\mathrm{y}-\mathrm{k}{)}^2}{{\mathrm{b}}^2}=1$
${\mathrm{a}}^2=4,{\mathrm{b}}^2=5$
a) Centre c = (h, k) = (-2,1)
b) Eccentricity $\left(\mathrm{e}\right)=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{4+5}{4}}=\frac{3}{2}$
c) Vertices $\mathrm{A}=(\mathrm{h}+\mathrm{a},\mathrm{k})=(-2+2,1)=(0,1)$
${\mathrm{A}}^{\mathrm{\prime }}=(\mathrm{h}-\mathrm{a},\mathrm{k})=(-2-2,1)=(-4,1)$
$\text{ d) foci }\mathrm{s}=(\mathrm{h}\pm \mathrm{ae},\mathrm{k})$
$=\left(-2\pm 2\cdot \frac{3}{2},1\right)=(-2\pm 3,1)=(1,1),(-5,1)$
e) Length of latus rectum = $\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2\times 5}{2}=5$
f) Equations of directrices
$\begin{array}{l}\mathrm{x}=\mathrm{h}\pm \frac{\mathrm{a}}{\mathrm{e}}=-2\pm \frac{2}{\frac{3}{2}}=-2\pm \frac{4}{3}=\frac{-6\pm 4}{3}\\ 3\mathrm{x}=-2,3\mathrm{x}=-10\\ 3\mathrm{x}+2=0,3\mathrm{x}+10=0\end{array}$
ii)
$\begin{array}{l}4{\mathrm{x}}^2-9{\mathrm{y}}^2-8\mathrm{x}-32=0\\ 4{\mathrm{x}}^2-8\mathrm{x}-9{\mathrm{y}}^2-32=0\\ 4\left({\mathrm{x}}^2-2\mathrm{x}\right)-9{\mathrm{y}}^2-32=0\\ 4\left({\mathrm{x}}^2-2\left(1\right)\mathrm{x}+1-1\right)-9{\mathrm{y}}^2-32=0\\ 4(\mathrm{x}-1{)}^2-4-9{\mathrm{y}}^2-32=0\\ 4(\mathrm{x}-1{)}^2-9{\mathrm{y}}^2=36\\ \frac{4(\mathrm{x}-1{)}^2}{36}-\frac{9{\mathrm{y}}^2}{36}=1\Rightarrow \frac{(\mathrm{x}-1{)}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1\\ {\mathrm{a}}^2=9,{\mathrm{b}}^2=4\end{array}$
a) Centre c = (h, k) = (1, 0)
$$\begin{gathered} \text{ b) Eccentricity }\mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{9+4}{9}}=\frac{\sqrt{13}}{3} \\ \text{ c) Vertices }\mathrm{A}=(\mathrm{h}+\mathrm{a},\mathrm{k})=(1+3,0)=(4,0) \\ {\mathrm{A}}^{\mathrm{\prime }}=(\mathrm{h}-\mathrm{a},\mathrm{k})=(1-3,0)=(-2,0) \\ \text{ d) foci }=(\mathrm{h}\pm \mathrm{ae},\mathrm{k})=\left(1\pm 3\times \frac{\sqrt{13}}{3},0\right)=(1\pm \sqrt{13},0) \\ \text{ e) Length of latusrectum }=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=2\times \frac{4}{3}=\frac{8}{3} \\ \text{ f) Equations of directrices }\mathrm{x}=\mathrm{h}\pm \frac{\mathrm{a}}{\mathrm{e}} \\ =1\pm \frac{3}{\frac{\sqrt{13}}{3}}=1\pm \frac{9}{\sqrt{13}} \end{gathered}$$
iii)
Sol: Equation of the hyperbola is
$\begin{array}{l}4{\mathrm{x}}^2-5{\mathrm{y}}^2-16\mathrm{x}+10\mathrm{y}+31=0\\ \Rightarrow 4\left({\mathrm{x}}^2-4\mathrm{x}\right)-5\left({\mathrm{y}}^2-2\mathrm{y}\right)=-31\\ \Rightarrow 4\left[{\mathrm{x}}^2-4\mathrm{x}+4\right]-5\left({\mathrm{y}}^2-2\mathrm{y}+1\right)=-31+16-5\\ \Rightarrow 4\left[(\mathrm{x}-2{)}^{2}5(\mathrm{y}+1{)}^2\right]=-20\\ \Rightarrow \frac{(\mathrm{y}-1{)}^2}{4}-\frac{(\mathrm{x}-2{)}^2}{5}=1\end{array}$
comparing with $\frac{(\mathrm{y}-\mathrm{K}{)}^2}{{\mathrm{b}}^2}-\frac{(\mathrm{x}-\mathrm{h}{)}^2}{{\mathrm{a}}^2}=1$
$$\begin{gathered} {\mathrm{a}}^2=5,{\mathrm{b}}^2=4,\mathrm{h}=2,\mathrm{k}=+1 \\ \text{ centre }=(\mathrm{h},\mathrm{k})=(2,1) \\ \mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{b}}^2}}=\sqrt{\frac{5+4}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2} \\ \text{ foci }=(\mathrm{h},\mathrm{K}\pm \mathrm{be})=(2,1\pm 3)=(2,4)\text{ and }(2,-2) \end{gathered}$$
equations of the directrices are
$\begin{array}{l}\mathrm{y}-\mathrm{K}=\pm \frac{\mathrm{b}}{\mathrm{e}}\Rightarrow \mathrm{y}-1=\pm 2\cdot \frac{2}{3}\\ \Rightarrow 3\mathrm{y}-3=\pm 4\Rightarrow 3\mathrm{y}-7=0\text{ and }3\mathrm{y}+1=0\end{array}$
iv)
Sol: Equation of the hyperbola is
$\begin{array}{l}5{\mathrm{x}}^2-4{\mathrm{y}}^2+20\mathrm{x}+8\mathrm{y}-4\\ \Rightarrow 5\left({\mathrm{x}}^2+4\mathrm{x}\right)-4\left({\mathrm{y}}^2-2\mathrm{y}\right)=4\\ \Rightarrow 5\left({\mathrm{x}}^2+4\mathrm{x}+4\right)-4\left({\mathrm{y}}^2-2\mathrm{y}+1\right)=4+20-4\\ \Rightarrow 5(\mathrm{x}+2{)}^2-4(\mathrm{y}-1{)}^2=20\\ \Rightarrow \frac{(\mathrm{x}+2{)}^2}{4}-\frac{(\mathrm{y}-1{)}^2}{5}=1\end{array}$
comparing with $\frac{(\mathrm{x}-\mathrm{h}{)}^2}{{\mathrm{a}}^2}-\frac{(\mathrm{y}-\mathrm{K}{)}^2}{{\mathrm{b}}^2}=1$
$$\begin{gathered} {\mathrm{a}}^2=4,{\mathrm{b}}^2=5,\mathrm{h}=-2,\mathrm{K}=1 \\ \text{ centre }=(\mathrm{h},\mathrm{K})=(-2,1) \\ \text{ foci }=(\mathrm{h}\pm \mathrm{ae},\mathrm{K}) \\ {\mathrm{a}}^2{\mathrm{e}}^2={\mathrm{a}}^2+{\mathrm{b}}^2=4+5=9;\mathrm{ae}=3\Rightarrow \mathrm{e}=\frac{\mathrm{ae}}{\mathrm{a}}=\frac{3}{2} \\ \text{ foci are }(-2\pm 3,1)=(1,1)\text{ and }(-5,1) \\ \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{directrices} \mathrm{are} \\ \mathrm{x}-\mathrm{h}=\pm \frac{\mathrm{a}}{\mathrm{e}}\Rightarrow \mathrm{x}+2=\pm 2\cdot \frac{2}{3}\Rightarrow 3\mathrm{x}+6=\pm 4 \\ \text{ i.e }3\mathrm{x}+6=4\text{ or }3\mathrm{x}+6=-4 \\ \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{directrices} \mathrm{are} 3\mathrm{x} + 2 = 0 \mathrm{and} 3\mathrm{x} + 10 = 0 \\ \text{ length of the latus rectum }=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2.5}{2}=5 \end{gathered}$$