A conic section in which the sum of the distances from the 2 foci to any point on it is a constant can be termed an ellipse. A hyperbola is obtained if an ellipse is turned inside out. In this article, you will learn how to find the parametric coordinates of hyperbola.
In the case of a hyperbola, the difference in the distance between the foci remains constant, and also, the 2 diagonal asymptotes cross at the centre. An auxiliary circle can be defined as the diameter of the circle described on the transverse axis of a hyperbola.
Standard Forms of a Hyperbola
The two standard forms of a hyperbola are
- Horizontal form: Centre is at the origin, and the hyperbola is symmetrical about the y-axis. The equation is x2 / a2 – y2 / b2 = 1. Here, the asymptotes of the hyperbola are y = [b / a]* x and y = [−b / a] * x.
- Vertical form: Centre is at the origin, and the hyperbola is symmetrical about the x-axis. The equation is y2 / a2 − x2 / b2 = 1 , where the asymptotes of the hyperbola are x = [b / a] * y and x = [−b / a] * y.
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How to Write the Parametric Form of a Hyperbola
The two ways to write the parametric form of a hyperbola are given by
A horizontal hyperbola
F (t) = (x (t), y (t))
x (t) = a sec (t)
y (t) = b tan (t)
A vertical hyperbola
F (t) = (x (t), y (t))
x (t) = b tan (t)
y (t) = a sec (t)
The parameter t lies in the interval 0 ≤ t < 2π. The angle ‘t’ is called the eccentric angle of the point [a sec (t), b tan (t)].
Solved Problems of Parametric Coordinates
Example 1: If the centre, vertex and focus of a hyperbola be (0, 0), (4, 0) and (6, 0), respectively, then what is the equation of the hyperbola?
Solution:
Centre (0, 0), vertex (4, 0) and focus (6, 0)
ae = 4
e = 3 / 2
Hence, the required equation is x2 / 16 − y2 / 20 = 1
i.e., 5x2 − 4y2 = 80
Example 2: The directrix of the hyperbola is x2 / 9 − y2 / 4 = 1 is __________.
Solution:
Directrix of hyperbola x = a / e, where e = √([b2 + a2] / [a2]) = √[b2 + a2] / [a]
Directrix is, x = a2 / √[a2 + b2]
=9 / √[9 + 4]
x = 9 / 13
Example 3: The equation of the hyperbola whose foci are (6, 4) and (4, 4) and eccentricity 2 is given by ____________.
Solution:
Foci are (6, 4) and (4, 4)
e = 2 and centre is ([6 − 4] / 2, 4) = (1, 4)
6 = 1 + ae
ae = 5
a = 5 / 2 and b = [5 / 2] * (√3)
Hence, the required equation is [(x − 1)2 / (25/4)] − [(y − 4)2 / (75 / 4)] = 1 or
12x2 − 4y2 − 24x + 32y − 127 = 0.
Example 4: The latus rectum of the hyperbola 9x2 − 16y2 − 18x − 32y − 151 = 0 is ________.
Solution:
9x2 − 18x + 9 − 16y2 − 32y − 16 = 144
⇒ (x − 1)2 / 16 − (y + 1)2 / 9 = 1
Latus rectum = 2b2/ a = [2 × 9] / [4] = 9 / 2
Example 5: The equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 1) and eccentricity = √3, is ___________.
Solution:
S (1, 1), directrix is 2x + y = 1 and e = √3
Now, let the various points be (h, k), then accordingly,
√[(h − 1)2 + (k − 1)2] / ([2h + k − 1] / √5) = √3
Squaring both the sides, we get, 5 [(h − 1)2 + (k − 1)2] = 3 (2h + k − 1)2
On simplification, the required locus is 7x2 + 12xy − 2y2 − 2x + 4y − 7 = 0
Example 6: The eccentricity of the hyperbola 5x2 − 4y2 + 20x + 8y = 4 is __________.
Solution:
Given equation of hyperbola is 5x2 − 4y2 + 20x + 8y = 4
5 (x + 2)2 − 4 (y − 1)2 = 20
(x + 2)2 / 4 − (y − 1)2 / 5 = 1
From b2 = a2 (e2 − 1),
5 = 4 (e2 − 1)
⇒ e2 = 9 / 4
⇒ e = 3 / 2
Example 7: The eccentricity of the hyperbola conjugate to x2 – 3y2 = 2x + 8 is
Solution:
Given, the equation of the hyperbola is
Conjugate of this hyperbola is
Here,
Example 8: If the foci of the ellipse
Solution:
Hyperbola is
Therefore, foci
Therefore, the focus of ellipse = (4e, 0), i.e., (3, 0)
Hence,
Example 9: If e and e’ are eccentricities of hyperbola and its conjugate, respectively, then
Solution:
Let hyperbola be
Then its conjugate will be,
If e is the eccentricity of the hyperbola (i), then
Similarly, if e’ is the eccentricity of conjugate (ii), then
Adding (iii) and (iv),
Example 10: What will be the equation of that chord of hyperbola 25x2 – 16y2 = 400, whose midpoint is (5, 3)?
Solution:
According to question,
Equation of required chord is
Here,
So from (i), the required chord is
Parametric Coordinates and Parametric Equations of Parabola
![Daily Quiz Parametric Coordinates and Parametric Equations of Parabola](https://cdn1.byjus.com/wp-content/uploads/2021/06/Parametric-Coordinates-and-Parametric-Equations-of-Parabola.jpg)
Frequently Asked Questions
What do you mean by hyperbola?
A hyperbola is the locus of all those points in a plane, such that the difference in their distances from two fixed points in the plane is a constant.
Give the parametric form of hyperbola.
The equations x = a sec θ, y = b tan θ gives the parametric equations of the hyperbola (x2/a2) – (y2/b2) = 1, where θ is the parameter.
What do you mean by a rectangular hyperbola?
A hyperbola having the major axis and the minor axis of equal length is known as a rectangular hyperbola.
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