Question

If $\mathrm{t}$ is a parameter then $\mathrm{x}=\mathrm{a}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)$ and $\mathrm{y}=\mathrm{b}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)$ represents

Answer 1

Explanation for the correct option:

Finding the relation between $\mathbf{x}$ and $\mathbf{y}$:

We need to eliminate $\mathrm{t}$ to find the relation between $\mathrm{x}$ and $\mathrm{y}$,

Given equation $\mathrm{x}=\mathrm{a}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)$ and $\mathrm{y}=\mathrm{b}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)$ can be rewritten as

$\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)...\left(\mathrm{i}\right)$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{b}}=\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)...\left(\mathrm{ii}\right)$

squaring equation $\left(\mathrm{i}\right)$, we get

$$\begin{gathered} {\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^2={\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)}^2 \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}={\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}+2...\left(\mathrm{iii}\right) \end{gathered}$$

squaring equation $\left(\mathrm{ii}\right)$, we get

$\begin{array}{rcl}{\left(\frac{\mathrm{y}}{\mathrm{b}}\right)}^2& =& {\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)}^2\\ & \Rightarrow & \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}={\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}-2...\left(\mathrm{iv}\right)\end{array}$

Substracting equation $\left(\mathrm{iv}\right)$ from $\left(\mathrm{iii}\right)$

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

$$\begin{gathered} \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=4 \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\left(2\mathrm{a}\right)}^2}-\frac{{\mathrm{y}}^2}{{\left(2\mathrm{b}\right)}^2}=1 \end{gathered}$$

which is the general equation of hyperbola $\begin{array}{rcl}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}& =& 1\end{array}$

Hence, the relation between $\mathbf{x}$ and $\mathbf{y}$ represents the general equation of hyperbola.