We will learn in the simplest way to find the parametric equations of the hyperbola.

The circle described as the diameter on the transverse axis of a hyperbola is called the auxiliary circle.

If ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 is a hyperbola, then is the auxiliary circle is x (^ {2} ) + y (^ {2} ) = a (^ {2} ).

The equation of the hyperbola is ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 The transverse axis of the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 is AA ‘ and its length = 2a. Obviously the equation of the circle described as diameter on AA ‘is x (^ {2} ) + y (^ {2} ) = a (^ {2} ) (since the center of the circle is the center C ( 0, 0) the hyperbola).

Therefore the equation of the auxiliary circle of the hyperbola is ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 is, x (^ {2} ) + y (^ {2} ) = a (^ {2} )

Let P (x, y) be an arbitrary point on the hyperbolic equation let ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ { 2}} ) = 1

Now draw from P from PM perpendicular to the transverse axis of the hyperbola. Take again a point Q on the auxiliary circle x (^ {2} ) + y (^ {2} ) = a (^ {2} ) with ∠CQM = 90 °.

Connect points C and Q. The length of QC = a. Let ∠MCQ = θ again. The angle ∠MCQ = θ is called the eccentric angle of point P on the hyperbola.

From the right-angled ∆CQM we now get

( frac {CQ} {MC} ) = cos

or a / MC = a / sec θ

or, MC = one second θ

Therefore, the abscissa of P = MC = x = a sec θ

Since the point P (x, y) on the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2} } ) = 1, hence

( frac {a ^ {2} sec ^ {2} θ} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1, (Since x = one second θ)

( frac {y ^ {2}} {b ^ {2}} ) = sec (^ {2} ) θ – 1

( frac {y ^ {2}} {b ^ {2}} ) = tan (^ {2} ) θ

y (^ {2} ) = b (^ {2} ) tan (^ {2} ) θ

y = btan θ

Therefore, the coordinates of P are (a sec θ, b tan θ).

Therefore, for all values ​​of θ, the point P (a sec θ, b tan θ) always lies on the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1

Thus, the coordinates of the point with the eccentric angle can be written as (a sec θ, b tan). Here, (a sec θ, b tan θ) are called parametric coordinates of the point P.

The equations x = a sec θ, y = b tan θ taken together are called the parametric equations of the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2 }} {b ^ {2}} ) = 1; where θ is the parameter (θ is called the eccentric angle of point P).

Solved example to find the parametric equations of a hyperbola:

1. Find the parametric coordinates of the point (8, 3√3) on the hyperbola 9x (^ {2} ) – 16y (^ {2} ) = 144.

Solution:

The given equation of the hyperbola is 9×2 – 16y2 = 144

⇒ ( frac {x ^ {2}} {16} ) – ( frac {y ^ {2}} {9} ) = 1

⇒ ( frac {x ^ {2}} {4 ^ {2}} ) – ( frac {y ^ {2}} {3 ^ {2}} ) = 1, which is the form of ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1.

Therefore,

a (^ {2} ) = 4 (^ {2} )

⇒ a = 4 and

b (^ {2} ) = 3 (^ {2} )

b = 3.

Hence we can assume the parametric coordinates of the point (8, 3√3) as (4 sec θ, 3 tan θ).

So we have 4 sec = 8

⇒ sec θ = 2

θ = 60 °

We know that for all values ​​of θ the point (a sec θ, b tan θ) always on the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac { y ^ {2}} {b ^ {2}} ) = 1

Therefore, (a sec θ, b tan θ) are known as the parametric coordinates of the point.

Therefore the parametric coordinates of the point are (8, 3√3) (4 sec. 60 °, 3 tan 60 °).

2. P (a sec θ, a tan θ) is a variable point on the hyperbola x (^ {2} ) – y (^ {2} ) = a (^ {2} ), and M ( 2a, 0) is a fixed point. Prove that the location of the center of AP is a rectangular hyperbola.

Solution:

Let (h, k) be the center of the line segment AM.

Therefore h = ( frac {a sec θ + 2a} {2} )

⇒ per second θ = 2 (h – a)

(one second θ) (^ {2} ) = [2(h – a)] (^ {2} ) …………………. (I)

and k = ( frac {a tan θ} {2} )

⇒ a tan θ = 2k

(a tan θ) (^ {2} ) = (2k) (^ {2} ) …………………. (ii)

If we now form (i) – (ii), we get

(one second θ) (^ {2} ) – (one tan θ) (^ {2} ) = [2(h – a)] (^ {2} ) – (2k) (^ {2} )

⇒ a (^ {2} ) (sec (^ {2} ) θ – tan (^ {2} ) θ) = 4 (h – a) (^ {2} ) – 4k (^ {2} )

⇒ (h – a) (^ {2} ) – k (^ {2} ) = ( frac {a ^ {2}} {4} ).

Therefore the equation for the locus of (h, k) (x – a) (^ {2} ) – y (^ {2} ) = ( frac {a ^ {2}} {4 } ), which is the equation of a rectangular hyperbola.

The hyperbole