Parametric Coordinates of Hyperbola

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.

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