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 $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1 is a hyperbola, then its auxiliary circle is x$^{2}$ + y$^{2}$ = a$^{2}$.

Let the equation of the hyperbola be, $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1

$Parametric Equation of the Hyperbola$

The transverse axis of the hyperbola $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1 is AA’ and its length = 2a. Clearly, the equation of the circle described on AA’ as diameter is x$^{2}$ + y$^{2}$ = a$^{2}$ (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 $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1 is, x$^{2}$ + y$^{2}$ = a$^{2}$

Let P (x, y) be any point on the equation of the hyperbola be $\frac{x^{2}}{a^{2}}$ -$\frac{y^{2}}{b^{2}}$ = 1

Now from P draw PM perpendicular to the transverse axis of the hyperbola. Again take a point Q on the auxiliary circle x$^{2}$ + y$^{2}$ = a$^{2}$ 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,

$\frac{CQ}{MC}$ = 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 $\frac{x^{2}}{a^{2}}$ -$\frac{y^{2}}{b^{2}}$ = 1 hence,

$\frac{a^{2}sec^{2} θ }{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1, (Since, x = a sec 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 = 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 $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 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 $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 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 9x$^{2}$ - 16y$^{2}$ = 144.

Solution:

The given equation of the hyperbola is 9x2 - 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.

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 $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 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 x$^{2}$ - y$^{2}$ = a$^{2}$, 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 = $\frac{a sec θ + 2a}{2}$

⇒ a sec θ = 2(h - a)

(a sec 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)

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

(a sec θ)$^{2}$ - (a 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$^{2}$ = $\frac{a^{2}}{4}$.