HyperbolaHardBloom L3
Question
Which of the following equations in parametric form can represent a hyperbola, where ' $t$ ' is a parameter ?
Options
A.$x = \frac{a}{2}\left( t + \frac{1}{t} \right)$ and $y = \frac{b}{2}\left( t - \frac{1}{t} \right)$
B.$\frac{tx}{a} - \frac{y}{b} + t = 0$ and $\frac{x}{a} + \frac{ty}{b} - 1 = 0$
C.$x = e^{t} + e^{- t}$ and $y = e^{t} - e^{- t}$
D.$x^{2} - 6 = 2cost$ and $y^{2} + 2 = 4\cos^{2}\frac{t}{2}$
Solution
Sol. a) $\left( \frac{x}{a} \right)^{2} - \left( \frac{y}{b} \right)^{2} = \frac{1}{4}\left\lbrack \left( t + \frac{1}{t} \right)^{2} - \left( t - \frac{1}{t} \right)^{2} \right\rbrack = 1$
c) $x^{2} - y^{2} = \left( e^{t} + e^{- t} \right)^{2} - \left( e^{t} - e^{- t} \right)^{2} = 4$
$$\begin{matrix} \text{~}\text{d)}\text{~} & x^{2} - 6 = 2cost \\ \Rightarrow & y^{2} + 2 = 2(1 + cost) \\ \Rightarrow & x^{2} - 6 = y^{2} \\ \Rightarrow & x^{2} - y^{2} = 6 \end{matrix}$$
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