Question

Which of the following equations in parametric form can represent a hyperbolic profile, where $t$ is a parameter.

• A. $x=\frac{a}{2}(t+\frac{1}{t})$ & $y=\frac{b}{2}(t-\frac{1}{t})$
• B. $\frac{tx}{a}-\frac{y}{b}+t=0$ & $\frac{x}{a}-\frac{ty}{b}-1=0$
• C. $x=e^t+e^{-t}$ & $x=e^t-e^{-t}$
• D. $x^2-6=2\cos t$ & $y^2+2=4{\cos }^2\frac{t}{2}$

Answer 1

The correct option is A $x=\frac{a}{2}(t+\frac{1}{t})$ & $y=\frac{b}{2}(t-\frac{1}{t})$

$\left(a\right)$ We have $x=\frac{a}{2}(t+\frac{1}{t})$ and $y=\frac{b}{2}(t-\frac{1}{t})$

$\Rightarrow \frac{2x}{a}=t+\frac{1}{t}$ and $\frac{2y}{b}=t-\frac{1}{t}$

$\Rightarrow {\left(\frac{2x}{a}\right)}^2={(t+\frac{1}{t})}^2$ and ${\left(\frac{2y}{b}\right)}^2={(t-\frac{1}{t})}^2$

$\Rightarrow \frac{4x^2}{a^2}=t^2+\frac{1}{t^2}+2$ and $\frac{4y^2}{b^2}=t^2+\frac{1}{t^2}-2$

$\Rightarrow \frac{4x^2}{a^2}-\frac{4y^2}{b^2}=t^2+\frac{1}{t^2}+2-t^2-\frac{1}{t^2}+2$

$\Rightarrow \frac{4x^2}{a^2}-\frac{4y^2}{b^2}=4$

$\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is in parametric form can represents a hyperbolic profile.

$\left(b\right)\frac{tx}{a}-\frac{y}{b}+t=0$ and $\frac{x}{a}-\frac{ty}{b}-1=0$

$\Rightarrow \frac{tx}{a}+t=\frac{y}{b}$ and $\frac{x}{a}-1=\frac{ty}{b}$

$\Rightarrow t(\frac{x}{a}+1)=\frac{y}{b}$ and $\frac{b}{y}(\frac{x}{a}-1)=t$

$\Rightarrow t\left(\frac{x+a}{a}\right)=\frac{y}{b}$ 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.

$\left(c\right)x=e^t+e^{-t}$ and $x=e^t-e^{-t}$

$\Rightarrow x^2=e^{2t}+e^{-2t}+2e^t{e}^{-t}$ and $x^2=e^{2t}+e^{-2t}-2e^t{e}^{-t}$

$\Rightarrow x^2=e^{2t}+e^{-2t}+2$ 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.

$\left(d\right)x^2-6=2\cos t$ and $y^2+2=4{\cos }^2\frac{t}{2}$

$\Rightarrow x^2=2\cos t+6$ and $y^2=4{\cos }^2\frac{t}{2}-2$

$\Rightarrow x^2-y^2=2\cos t+6-4{\cos }^2\frac{t}{2}+2$

$\Rightarrow x^2-y^2=2(2{\cos }^2\frac{t}{2}-1)+8-4{\cos }^2\frac{t}{2}$

$\Rightarrow x^2-y^2=4{\cos }^2\frac{t}{2}-2+8-4{\cos }^2\frac{t}{2}$

$\Rightarrow x^2-y^2=6$ is in parametric form can represents a hyperbolic profile.