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Parametric Equation of the Hyperbola

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

The circle described on the transverse axis of a hyperbola as diameter is called its Auxiliary Circle.

If x2a2 - y2b2 = 1 is a hyperbola, then its auxiliary circle is x2 + y2 = a2.

Let the equation of the hyperbola be, x2a2 - y2b2 = 1  

Parametric Equation of the Hyperbola

The transverse axis of the hyperbola x2a2 - y2b2 = 1 is AA’ and its length = 2a. Clearly, the equation of the circle described on AA’ as diameter is x2 + y2 = a2 (since the centre of the circle is the centre C (0, 0) of the hyperbola).

Therefore, the equation of the auxiliary circle of the hyperbola x2a2 - y2b2 = 1 is, x2 + y2 = a2

Let P (x, y) be any point on the equation of the hyperbola be x2a2 -y2b2 = 1

Now from P draw PM perpendicular to the transverse axis of the hyperbola. Again take a point Q on the auxiliary circle x2 + y2 = a2 such that ∠CQM = 90°.

Join the point C and Q. The length of QC = a. Again, let ∠MCQ = θ. The angle ∠MCQ = θ is called the eccentric angle of the point P on the hyperbola.

Now from the right-angled  ∆CQM we get,

CQMC = cos θ          

or, a/MC  =   a/sec θ       

or, MC  = a sec θ

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

Since the point P (x, y) lies on the hyperbola x2a2 -y2b2 = 1 hence,

a2sec2θa2 - y2b2 = 1, (Since, x = a sec θ)

y2b2 = sec2 θ – 1

y2b2 = tan2 θ

y2 = b2 tan2 θ

y = b tan θ

Hence, the co-ordinates 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 x2a2 - y2b2  = 1  

Thus, the co-ordinates of the point having eccentric angle θ can be written as (a sec θ, b tan θ). Here (a sec θ, b tan θ) are known as the parametric co-ordinates of the point P.

The equations x = a sec θ, y = b tan θ taken together are called the parametric equations of the hyperbola x2a2 - y2b2 = 1; where θ is parameter (θ is called the eccentric angle of the point P).


Solved example to find the parametric equations of a hyperbola:

1. Find the parametric co-ordinates of the point (8, 3√3) on the hyperbola 9x2 - 16y2 = 144.

Solution:     

The given equation of the hyperbola is 9x2 - 16y2 = 144

x216 - y29 = 1

x242 - y232 = 1, which is the form of x2a2 - y2b2 = 1.  

Therefore,

a2 = 42 

⇒ a = 4 and   

b2 = 32     

⇒ b = 3.

Therefore, we can take the parametric co-ordinates of the point (8, 3√3) as (4 sec θ, 3 tan θ).

Thus we have, 4 sec θ = 8      

⇒ sec θ = 2        

⇒ θ = 60°

We know that for all values of θ the point (a sec θ, b tan θ) always lies on the hyperbola x2a2 - y2b2  = 1  

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

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

 

2. P (a sec θ, a tan θ) is a variable point on the hyperbola x2 - y2 = a2, and M (2a, 0) is a fixed point. Prove that the locus of the middle point of AP is a rectangular hyperbola.

Solution:        

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

Therefore, h = asecθ+2a2   

⇒ a sec θ = 2(h - a)

(a sec θ)2 = [2(h - a)]2 …………………. (i)

and k = atanθ2

⇒ a tan θ = 2k

(a tan θ)2 = (2k)2 …………………. (ii)

Now form (i) - (ii), we get,

(a sec θ)2 - (a tan θ)2 = [2(h - a)]2 - (2k)2

⇒ a2(sec2 θ - tan2 θ) = 4(h - a)2 - 4k2

⇒ (h - a)2 - k2 = a24.

Therefore, the equation to the locus of (h, k) is (x - a)2 - y2 = a24, which is the equation of a rectangular hyperbola.

The Hyperbola





11 and 12 Grade Math 

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