Due to length restrictions, we kindly invite to check the explanation herein for further details of the hyperbola.
How to analyze an hyperbola
Herein we have an hyperbola whose axis of symmetry is parallel to the y-axis and the major semiaxis length is in the y-direction. By analytical geometry, we know that eccentricities of hyperbolae are greater than 1.
a) The formula for eccentricity is:
e = √(a² + b²) / a (1)
Where:
- a - Major semiaxis length
- b - Minor semiaxis length
If we know that a = 4 and b = 3, then the eccentricity of the hyperbola is:
e = √(4² + 3²) / 4
e = 5 / 4
b) The coordinates of the two vertices of the hyperbola are:
V(x, y) = (h, k ± a) (2)
Where (h, k) are the coordinates of the center of the hyperbola.
V₁ (x, y) = (0, 4), V₂ (x, y) = (0, - 4)
The coordinates of the foci of the hyperbola are:
F(x, y) = (h, k ± c), where c = √(a² + b²). (3)
c = √(4² + 3²)
c = 5
F₁ (x, y) = (0, 5), F₂ (x, y) = (0, - 5)
The equations of the asymptotes of the hyperbola are:
y = ± (a / b) · x
y = ± (4 / 3) · x (4)
And the equations of the directrices of the hyperbola are:
y = k ± (2 · a - c)
y = 0 ± (8 - 5)
y = ± 3 (5)
The graph is presented in the image attached below.
c) The parametric equations for the hyperbola are the following formulae:
y = ± a · cosh t → y = ± 4 · cosh t (6)
x = b · sinh t → x = 3 · sinh t (7)
d) First, we determine the slopes of the two tangent lines by implicit differentiation:
m = (16 · x) / (9 · y)
m = (16 · 2.3) / [9 · (± 4.807)]
m = ± 0.851
Second, we find the intercept of each tangent line:
(x, y) = (2, 4.807)
b = 4.807 - 0.851 · 2
b = 3.105
y = 0.851 · x + 3.105 (8)
(x, y) = (2, - 4.807)
b = - 4.807 - (- 0.851) · 2
b = - 3.105
y = - 0.851 · x - 3.105 (9)
e) The definite integral of the arc length of the hyperbola is presented below:
[tex]s = \int\limits^{2}_{1} {\sqrt{\left(\frac{dx}{dt} \right)^{2}+\left(\frac{dy}{dt} \right)^{2}}} \, dt[/tex]
If we know that dx / dt = a² · sinh² t and dy / dt = b² · cosh² t, then the definite integral for the arc length is:
[tex]s = \int\limits^2_1 {\sqrt{a^{2}\cdot \sinh ^{2}t +b^{2}\cdot \cosh^{2}t}} \, dt[/tex] (10)
f) We apply the following substitutions on (1): x = r · cos θ, y = r · sin θ. Then, we have the polar form by algebraic handling:
r(θ) = (a · b) / (b² · sin² θ - a² · cos² θ) (11)
To learn more on hyperbolae: https://brainly.com/question/12919612
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