# 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

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,

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$$^{2}$$ = $$\frac{a^{2}}{4}$$.

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

The Hyperbola

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