$(a)$ We have $x=\dfrac{a}{2}\left(t+\dfrac{1}{t}\right)$ and $y=\dfrac{b}{2}\left(t-\dfrac{1}{t}\right)$
$\Rightarrow\,\dfrac{2x}{a}=t+\dfrac{1}{t}$ and $\dfrac{2y}{b}=t-\dfrac{1}{t}$
$\Rightarrow\,{\left(\dfrac{2x}{a}\right)}^{2}={\left(t+\dfrac{1}{t}\right)}^{2}$ and ${\left(\dfrac{2y}{b}\right)}^{2}={\left(t-\dfrac{1}{t}\right)}^{2}$
$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}+2$ and $\dfrac{4{y}^{2}}{{b}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}-2$
$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}-\dfrac{4{y}^{2}}{{b}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}+2-{t}^{2}-\dfrac{1}{{t}^{2}}+2$
$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}-\dfrac{4{y}^{2}}{{b}^{2}}=4$
$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ is in parametric form can represents a hyperbolic profile.
$(b)\dfrac{tx}{a}-\dfrac{y}{b}+t=0$ and $\dfrac{x}{a}-\dfrac{ty}{b}-1=0$
$\Rightarrow\,\dfrac{tx}{a}+t=\dfrac{y}{b}$ and $\dfrac{x}{a}-1=\dfrac{ty}{b}$
$\Rightarrow\,t\left(\dfrac{x}{a}+1\right)=\dfrac{y}{b}$ and $\dfrac{b}{y}\left(\dfrac{x}{a}-1\right)=t$
$\Rightarrow\,t\left(\dfrac{x+a}{a}\right)=\dfrac{y}{b}$ and $t=\dfrac{b}{y}\left(\dfrac{x-a}{a}\right)$
$\Rightarrow\,\dfrac{b}{y}\left(\dfrac{x-a}{a}\right)\left(\dfrac{x+a}{a}\right)=\dfrac{y}{b}$
$\Rightarrow\,\dfrac{b}{y}\left(\dfrac{{x}^{2}-{a}^{2}}{{a}^{2}}\right)=\dfrac{y}{b}$
$\Rightarrow\,\dfrac{{x}^{2}-{a}^{2}}{{a}^{2}}=\dfrac{{y}^{2}}{{b}^{2}}$
$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-1=\dfrac{{y}^{2}}{{b}^{2}}$
$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ is in parametric form can represents a hyperbolic profile.
$(c)x={e}^{t}+{e}^{-t}$ and $x={e}^{t}-{e}^{-t}$
$\Rightarrow\,{x}^{2}={e}^{2t}+{e}^{-2t}+2{e}^{t}{e}^{-t}$ and $ {x}^{2}={e}^{2t}+{e}^{-2t}-2{e}^{t}{e}^{-t}$
$\Rightarrow\,{x}^{2}={e}^{2t}+{e}^{-2t}+2$ and ${x}^{2}={e}^{2t}+{e}^{-2t}-2$
$\Rightarrow\,{x}^{2}-{x}^{2}={e}^{2t}+{e}^{-2t}+2-{e}^{2t}-{e}^{-2t}+2=4$ is not in parametric form can represents a hyperbolic profile.
$(d){x}^{2}-6=2\cos{t}$ and ${y}^{2}+2=4{\cos}^{2}{\dfrac{t}{2}}$
$\Rightarrow\,{x}^{2}=2\cos{t}+6$ and ${y}^{2}=4{\cos}^{2}{\dfrac{t}{2}}-2$
$\Rightarrow\,{x}^{2}-{y}^{2}=2\cos{t}+6-4{\cos}^{2}{\dfrac{t}{2}}+2$
$\Rightarrow\,{x}^{2}-{y}^{2}=2\left(2{\cos}^{2}{\dfrac{t}{2}}-1\right)+8-4{\cos}^{2}{\dfrac{t}{2}}$
$\Rightarrow\,{x}^{2}-{y}^{2}=4{\cos}^{2}{\dfrac{t}{2}}-2+8-4{\cos}^{2}{\dfrac{t}{2}}$
$\Rightarrow\,{x}^{2}-{y}^{2}=6$ is in parametric form can represents a hyperbolic profile.