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 rightangled ∆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 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 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 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 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 coordinates 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 coordinates 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 coordinates of the point.
Therefore, the parametric coordinates 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
11 and 12 Grade Math
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