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Which of the following equations in parametric form can represent a hyperbola, where $'t'$ is a parameter.
$x =$ $\frac{a}{2}$$\left( {t\,\, + \,\,\frac{1}{t}} \right)$ $\&$ $y = \frac{b}{2}$$\left( {t\,\, - \,\,\frac{1}{t}} \right)$
$x^2 - 6 = 2 cos t \,\,\&\,\, y^2 + 2 = 4 cos^2\frac{t}{2}$
$x = e^t + e^{-t} \,\,\& \,\,y = e^t -e^{-t}$
all of the above
Solution
(a) We have $x=\frac{a}{2}\left(t+\frac{1}{t}\right)$ and $y=\frac{b}{2}\left(t-\frac{1}{t}\right)$
$\Rightarrow \frac{2 x}{a}=t+\frac{1}{t}$ and $\frac{2 y}{b}=t-\frac{1}{t}$
$\Rightarrow\left(\frac{2 x}{a}\right)^{2}=\left(t+\frac{1}{t}\right)^{2}$ and $\left(\frac{2 y}{b}\right)^{2}=\left(t-\frac{1}{t}\right)^{2}$
$\Rightarrow \frac{4 x^{2}}{a^{2}}=t^{2}+\frac{1}{t^{2}}+2$ and $\frac{4 y^{2}}{b^{2}}=t^{2}+\frac{1}{t^{2}}-2$
$\Rightarrow \frac{4 x^{2}}{a^{2}}-\frac{4 y^{2}}{b^{2}}=t^{2}+\frac{1}{t^{2}}+2-t^{2}-\frac{1}{t^{2}}+2$
$\Rightarrow \frac{4 x^{2}}{a^{2}}-\frac{4 y^{2}}{b^{2}}=4$
$\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is in parametric form can represents a hyperbolic profile.
(b) $\frac{t x}{a}-\frac{y}{b}+t=0$ and $\frac{x}{a}-\frac{t y}{b}-1=0$
$\Rightarrow \frac{t x}{a}+t=\frac{y}{b}$ and $\frac{x}{a}-1=\frac{t y}{b}$
$\Rightarrow t\left(\frac{x}{a}+1\right)=\frac{y}{b}$ and $\frac{b}{y}\left(\frac{x}{a}-1\right)=t$
$\Rightarrow t\left(\frac{x+a}{a}\right)=\frac{y}{b}$ and $t=\frac{b}{y}\left(\frac{x-a}{a}\right)$
$\Rightarrow \frac{b}{y}\left(\frac{x-a}{a}\right)\left(\frac{x+a}{a}\right)=\frac{y}{b}$
$\Rightarrow \frac{b}{y}\left(\frac{x^{2}-a^{2}}{a^{2}}\right)=\frac{y}{b}$
$\Rightarrow \frac{x^{2}-a^{2}}{a^{2}}=\frac{y^{2}}{b^{2}}$
$\Rightarrow \frac{x^{2}}{a^{2}}-1=\frac{y^{2}}{b^{2}}$
$\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is in parametric form can represents a hyperbolic profile.
$(c) x=e^{t}+e^{-t}$ and $x=e^{t}-e^{-t}$
$\Rightarrow x^{2}=e^{2 t}+e^{-2 t}+2 e^{t} e^{-t}$ and $x^{2}=e^{2 t}+e^{-2 t}-2 e^{t} e^{-t}$
$\Rightarrow x^{2}=e^{2 t}+e^{-2 t}+2$ and $x^{2}=e^{2 t}+e^{-2 t}-2$
$\Rightarrow x^{2}-x^{2}=e^{2 t}+e^{-2 t}+2-e^{2 t}-e^{-2 t}+2=4$ is not in parametric form can represents a
hyperbolic profile.
$(d) x^{2}-6=2 \cos t$ and $y^{2}+2=4 \cos ^{2} \frac{t}{2}$
$\Rightarrow x^{2}=2 \cos t+6$ and $y^{2}=4 \cos ^{2} \frac{t}{2}-2$
$\Rightarrow x^{2}-y^{2}=2 \cos t+6-4 \cos ^{2} \frac{t}{2}+2$
$\Rightarrow x^{2}-y^{2}=2\left(2 \cos ^{2} \frac{t}{2}-1\right)+8-4 \cos ^{2} \frac{t}{2}$
$\Rightarrow x^{2}-y^{2}=4 \cos ^{2} \frac{t}{2}-2+8-4 \cos ^{2} \frac{t}{2}$
$\Rightarrow x^{2}-y^{2}=6$ is in parametric form can represents a hyperbolic profile.
