Question

Q. Find parametrisations of following level curves
Y^2-x^2=1, x^2/4+y^2/9 =1

Answer 1

Answer on Question #70823 – Math – Geometry

Question

Find parametrisations of following level curves $y^{2} - x^{2} = 1$, $x^2 / 4 + y^2 / 9 = 1$

Solution

1. Find a parametrisation of the hyperbola $y^{2} - x^{2} = 1$

$$(y - x) (y + x) = 1$$ $$\left\{ \begin{array}{l} y + x = t \\ y - x = \frac {1}{t} \end{array} \right.$$ $$\left\{ \begin{array}{l} y + x + y - x = t + \frac {1}{t} \\ y + x - (y - x) = t - \frac {1}{t} \end{array} \right.$$ $$\left\{ \begin{array}{l} 2 y = t + \frac {1}{t} \\ 2 x = t - \frac {1}{t} \end{array} \right.$$

Hence

$$\left\{ \begin{array}{l} y = \frac {1}{2} \left(t + \frac {1}{t}\right) \\ x = \frac {1}{2} \left(t - \frac {1}{t}\right) \end{array} \right.$$

is a parametrisation of the hyperbola $y^{2} - x^{2} = 1$.

2. Find a parametrisation of the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$

$$a = 2; b = 3;$$ $$\left\{ \begin{array}{l} \frac {x}{2} = \cos t \\ \frac {y}{3} = \sin t \end{array} \right.$$

Hence

$$\left\{ \begin{array}{l} x = 2 \cos t \\ y = 3 \sin t \end{array} \right.$$

is a parametrisation of the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$

Answer:

$$\left\{ \begin{array}{l} y = \frac {1}{2} \left(t + \frac {1}{t}\right) \\ x = \frac {1}{2} \left(t - \frac {1}{t}\right) \end{array} \right. \text{ is a parametrisation of the hyperbola } y ^ {2} - x ^ {2} = 1;$$ $$\left\{ \begin{array}{l} x = 2 \cos t \\ y = 3 \sin t \end{array} \right. \text{ is a parametrisation of the ellipse } \frac {x ^ {2}}{4} + \frac {y ^ {2}}{9} = 1.$$

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