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 #70822 – Math – Geometry

Question

1. Find parametrization of following level curve $y^2 - x^2 = 1$.

Solution

The equation of the conjugate hyperbola in Cartesian coordinates is given by

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

A parametrization of the conjugate hyperbola is

$$x = a \cdot \sinh t, \quad y = b \cdot \cosh t, \quad t \in R$$

(the hyperbolic identity $\cosh^2(t) - \sinh^2(t) = 1$ was applied),

then the curve

$$y^2 - x^2 = 1, \quad a = 1, \quad b = 1$$

has the following parametrization:

$$x = \sinh t, \quad y = \cosh t, \quad t \in R$$

**Answer**: $x = \sinh t$, $y = \cosh t$, $t \in R$

Question

2. Find parametrization of following level curve $x^2 / 4 + y^2 / 9 = 1$.

Solution

The equation of an ellipse in Cartesian coordinates is given by

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

A parametrization of ellipse curve is

$$x = a \cdot \cos t, \quad y = b \cdot \sin t, \quad 0 \leq t \leq 2\pi$$

(the trigonometric identity $\cos^2(t) + \sin^2(t) = 1$ was applied),

then the curve

$$\frac{x^2}{4} + \frac{y^2}{9} = 1, \quad a = 2, \quad b = 3$$

has the following parametrization:

$$x = 2 \cdot \cos t, \ y = 3 \cdot \sin t, \ 0 \leq t \leq 2 \pi$$

Answer: $x = 2 \cdot \cos t$, $y = 3 \cdot \sin t$, $0 \leq t \leq 2\pi$.

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