**Example 1:**

Find the center, vertices, axis, foci, eccentricity and asymptotes of the hyperbola $x^{2}\ –\ 4y^{2}\ +\ 2x\ –\ 16y\ –\ 19$ = $0$

**Solution: **

Arrange the given problem in the standard form $\frac{(x – h)^{2}}{a^{2}}$ – $\frac{(y – k)^{2}}{b^{2}}$ = $1$

To get that we first take the loose number -$19$ on the other side of the equal to $x^{2}\ –\ 4y^{2}\ +\ 2x\ –\ 16y$ = $19$

Arrange the $x$ terms and $y$ terms together $x^{2}\ +\ 2x\ –\ 4y^{2}\ –\ 16y$ = $19$

Factor the coefficient of the $x^{2}$ from the $x$ terms and similarly factor out the coefficient of the $y^{2}$ from the $y$ terms $1(x^{2}\ +\ 2x)\ –\ 4(y^{2}\ +\ 4y)$ = $19$

Complete the square within the parenthesis $1(x^{2}\ +\ 2x + 1)\ -\ 4(y^{2}\ +\ 4y + 4)$ = $19$ = $19\ +\ 1\ –\ 16$

or $ (x + 1)^{2}\ –\ 4(y + 2)^{2}$ = $4$

Divide throughout by the loose number $4$ present on the right side of the equal to

$\frac{(x + 1)^{2}}{4}$ – $\frac{4(y + 2)^{2}}{4}$ = $\frac{4}{4}$

$\frac{(x + 1)^{2}}{2^{2}}$ – $\frac{(y + 2)^{2}}{1^{2}}$ = $1$

Center : $(-1,\ -2)$

Vertices :$(h\ +\ a,\ k)$ and $(h\ –\ a,\ k)$ = $(-1\ +2,\ -2)$ and $(-1-2,\ -2)$ = $(1, -2),$ $(-3, -2)$

Axes : $a$ = $2$, $b$ = $1$

Foci :$(h + c, k)$ and $(h – c, k)\ where \ c$ = $\sqrt{(a^{2}\ +\ b^{2})}$ = $\sqrt{5}$.

Therefore foci = $(-1\ +\ \sqrt{5},\ -2)$ and $(-1\ - \sqrt{5},\ -2)$

Eccentricity = $\frac{c}{a}$ = $\frac{\sqrt{5}}{2}$

Asymptotes $y$ = $\pm$ $\frac{b}{a}$ $(x - h)\ +\ k$

or $y$ = $\pm$$\frac{1}{2}$ $(x\ +\ 1)$ - $2$