Step 1

a)For the given equation simplify it and compare with standard forms of conic sections to determine which conic section or a figure it represents as. ${x}^{2}\text{}-\text{}{y}^{2}\text{}-\text{}4x\text{}-\text{}6y\text{}+\text{}1=0$

${x}^{2}\text{}-\text{}4x\text{}+\text{}4\text{}-\text{}4\text{}-\text{}{y}^{2}\text{}-\text{}6y\text{}+\text{}9\text{}-\text{}9\text{}+\text{}1=0$

$({x}^{2}\text{}-\text{}4x\text{}+\text{}4)\text{}-\text{}({y}^{2}\text{}+\text{}6y\text{}+\text{}9)\text{}+\text{}1\text{}+\text{}9\text{}-\text{}4=0$

$(x\text{}-\text{}2{)}^{2}\text{}-\text{}(y\text{}+\text{}3)62\text{}+\text{}6=0$

$(y\text{}+\text{}3{)}^{2}\text{}-\text{}(x\text{}-\text{}2{)}^{2}=6$

$\frac{(y\text{}+\text{}3{)}^{2}}{6}\text{}-\text{}\frac{(x\text{}-\text{}2{)}^{2}}{6}=1$

Which is a hyperbola.

Step 2

Now to find the features of the given hyperbola there are standard forms where these features can be determined. Let us determine centre ‘C’ of this hyperbola by comparing with standard form of such hyperbola as. $\frac{(y\text{}-\text{}k{)}^{2}}{{a}^{2}}\text{}-\text{}\frac{(x\text{}-\text{}h{)}^{2}}{{b}^{2}}=1$

$\frac{(y\text{}+\text{}3{)}^{2}}{6}\text{}-\text{}\frac{(x\text{}-\text{}2{)}^{2}}{6}=1$

$C=(h,\text{}k)$ here $C=(2,\text{}-3)$

Step 3

Vertices "V" for this hyperbola can be determined to be as. $V=(h,\text{}k\text{}\pm \text{}a)$

${a}^{2}=6$

$a=\text{}\sqrt{6}$

$V=(2{\textstyle \phantom{\rule{0.167em}{0ex}}}-3\text{}\pm \text{}\sqrt{6})$

Step 4

Foci "F" for this hyperbola can be determined to be as $F=(h,\text{}k\text{}\pm \text{}c)$

${c}^{2}={a}^{2}\text{}+\text{}{b}^{2}$

${c}^{2}=6\text{}+\text{}6$

$c=\text{}\sqrt{12}$

$F=(2,\text{}-3\text{}\pm \text{}\sqrt{12})$

Step 5

Equation of asymptotes for this hyperbola can be determined to be as. $y=\text{}\pm \text{}\frac{a}{b}(x\text{}-\text{}h)\text{}+\text{}k$

$y=\text{}\pm \text{}\frac{\sqrt{6}}{\sqrt{6}}(x\text{}-\text{}2)\text{}-\text{}3$

$y=\text{}\pm \text{}(x\text{}-\text{}2)\text{}-\text{}3$

Step 6

Eccentricity "e" is calculated by the formula as. $e=\text{}\frac{\sqrt{{a}^{2}={b}^{2}}}{a}$

$=\text{}\frac{\sqrt{6\text{}+\text{}6}}{\sqrt{6}}$

$=\text{}\frac{\sqrt{12}}{\sqrt{6}}$

$=\text{}\frac{\sqrt{2}\text{}\sqrt{6}}{\sqrt{6}}$

$=\text{}\sqrt{2}$