# Parametric Coordinates of Hyperbola

A conic section in which the sum of the distances from the 2 foci to any point on it is a constant can be termed as an ellipse. A hyperbola is obtained if an ellipse is turned inside out. In this article, you will learn how to find the Parametric Coordinates of Hyperbola.

In the case of a hyperbola, the difference in the distance between the foci remains constant and also the 2 diagonal asymptotes cross at the centre. An auxiliary circle can be defined as the diameter of the circle described on the transverse axis of a hyperbola.

## Standard Forms of a Hyperbola

The two standard forms of a hyperbola are:

• Horizontal form: Center is at the origin and hyperbola is symmetrical about the y-axis. The equation is x2 / a2 – y2 / b2 = 1. Here, the asymptotes of the hyperbola are y = [b / a]* x and y = [−b / a] * x.
• Vertical form: Center is at the origin and hyperbola is symmetrical about the x-axis. The equation is y2 / a2 − x2 / b2 = 1 , where the asymptotes of the hyperbola are x = [b / a] * y and x = [−b / a] * y.

## How to write the Parametric Form of a Hyperbola?

The two ways to write the parametric form of a hyperbola are given by:

A horizontal hyperbola:

F (t) = (x (t), y (t))

x (t) = a sec (t)

y (t) = b tan (t)

A vertical hyperbola:

F (t) = (x (t), y (t))

x (t) = b tan (t)

y (t) = a sec (t)

The parameter t lies in the interval 0 ≤ t < 2π. The angle ‘t’ is called the eccentric angle of the point [a sec (t), b tan (t)].

## Solved Problems On Parametric Coordinates

Example 1: If the centre, vertex and focus of a hyperbola be (0, 0), (4, 0) and (6, 0) respectively, then what is the equation of the hyperbola?

Solution:

Centre (0, 0), vertex (4, 0) and focus (6, 0)

ae = 4

e = 3 / 2.

Hence, the required equation is x2 / 16 − y2 / 20 = 1

i.e., 5x2 − 4y2 = 80.

Example 2: The directrix of the hyperbola is x2 / 9 − y2 / 4 = 1 is __________.

Solution:

Directrix of hyperbola x = a / e, where e = √([b2 + a2] / [a2]) = √[b2 + a2] / [a]

Directrix is, x = a2 / √[a2 + b2]

=9 / √[9 + 4]

x = 9 / 13

Example 3: The equation of the hyperbola whose foci are (6, 4) and (4, 4) and eccentricity 2 is given by ____________.

Solution:

Foci are (6, 4) and (4, 4)

e = 2 and centre is ([6 − 4] / 2, 4) = (1, 4)

6 = 1 + ae

ae = 5

a = 5 / 2 and b = [5 / 2] * (√3)

Hence, the required equation is [(x − 1)2 / (25/4)] − [(y − 4)2 / (75 / 4)] = 1 or

12x2 − 4y2 − 24x + 32y − 127 = 0.

Example 4: The latus rectum of the hyperbola 9x2 − 16y2 − 18x − 32y − 151 = 0 is ________.

Solution:

9x2 − 18x + 9 − 16y2 − 32y − 16 = 144

⇒ (x − 1)2 / 16 − (y + 1)2 / 9 = 1

Latus rectum = 2b2/ a = [2 × 9] / [4] = 9 / 2

Example 5: The equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 1) and eccentricity = √3, is ___________.

Solution:

S (1, 1), directrix is 2x + y = 1 and e = √3

Now let the various points be (h, k), then accordingly

√[(h − 1)2 + (k − 1)2] / ([2h + k − 1] / √5) = √3

Squaring both the sides, we get 5 [(h − 1)2 + (k − 1)2] = 3 (2h + k − 1)2

On simplification, the required locus is 7x2 + 12xy − 2y2 − 2x + 4y − 7 = 0.

Example 6: The eccentricity of the hyperbola 5x2 − 4y2 + 20x + 8y = 4 is __________.

Solution:

Given equation of hyperbola is 5x2 − 4y2 + 20x + 8y = 4

5 (x + 2)2 − 4 (y − 1)2 = 20

(x + 2)2 / 4 − (y − 1)2 / 5 = 1

From b2 = a2 (e2 − 1),

5 = 4 (e2 − 1)

⇒ e2 = 9 / 4

⇒ e = 3 / 2

Example 7: The eccentricity of the hyperbola conjugate to ${{x}^{2}}-3{{y}^{2}}=2x+8$ is

Solution:

Given, equation of hyperbola is

${{x}^{2}}-3{{y}^{2}}=2x+8 \\ {{x}^{2}}-2x-3{{y}^{2}}=8\\ {{(x-1)}^{2}}-3{{y}^{2}}=9\\ \frac{{{(x-1)}^{2}}}{9}-\frac{{{y}^{2}}}{3}=1$

Conjugate of this hyperbola is $-\frac{{{(x-1)}^{2}}}{9}+\frac{{{y}^{2}}}{3}=1$ and its eccentricity $(e)=\sqrt{\left( \frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} \right)}$

Here,

${{a}^{2}}=9\\ {{b}^{2}}=3\\ e=\sqrt{\frac{9+3}{3}}=2.$

Example 8: If the foci of the ellipse $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ and the hyperbola $\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}$ coincide, then find the value of ${{b}^{2}}$.

Solution:

Hyperbola is

$\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\\ a=\sqrt{\frac{144}{25}},\,\,b=\sqrt{\frac{81}{25}},\,\,{{e}_{1}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}$

Therefore, foci $=(a{{e}_{1}},0)=\left( \frac{12}{5}.\frac{5}{4},0 \right)=(3,\,0)$

Therefore, focus of ellipse $=(4e,0) \text \ i.e. \ (3,\,0)$

Hence ${{b}^{2}}=16\left( 1-\frac{9}{16} \right)=7.$

Example 9: If e and e’ are eccentricities of hyperbola and its conjugate respectively, then

$A) {{\left( \frac{1}{e} \right)}^{2}}+{{\left( \frac{1}{e’} \right)}^{2}}=1\\ B) \frac{1}{e}+\frac{1}{e’}=1\\ C) {{\left( \frac{1}{e} \right)}^{2}}+{{\left( \frac{1}{e’} \right)}^{2}}=0\\ D) \frac{1}{e}+\frac{1}{e’}=2\\$

Solution:

Let hyperbola be $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1 \rightarrow (i)$

Then its conjugate will be, $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1 \rightarrow (ii)$

If e is eccentricity of hyperbola (i), then

${{b}^{2}}={{a}^{2}}({{e}^{2}}-1) \\ \frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{({{a}^{2}}+{{b}^{2}})} \rightarrow (iii)$

Similarly if e’ is eccentricity of conjugate (ii), then

${{a}^{2}}={{b}^{2}}(e{{‘}^{2}}-1)\\ \frac{1}{e{{‘}^{2}}}=\frac{{{b}^{2}}}{({{a}^{2}}+{{b}^{2}})} \rightarrow (iv)$

$\frac{1}{{{(e’)}^{2}}}+\frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1.$

Example 10: What will be equation of that chord of hyperbola $25{{x}^{2}}-16{{y}^{2}}=400,$ whose mid point is (5, 3)?

Solution:

According to question, $S\equiv \,25{{x}^{2}}-16{{y}^{2}}-400=0$

Equation of required chord is ${{S}_{1}}=T \rightarrow (i)$

Here,

${{S}_{1}}=25{{(5)}^{2}}-16{{(3)}^{2}}-400 \\ =625-144-400=81$ and

$T\equiv 25x{{x}_{1}}-16y{{y}_{1}}-400,$ where

${{x}_{1}}=5,\,{{y}_{1}}=3\\ =25(x)(5)-16(y)(3)-400\\ =125x-48y-400$

So from (i), required chord is

$125x-48y-400=81\\ 125x-48y=481.$