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 x23y2=2x+8{{x}^{2}}-3{{y}^{2}}=2x+8 is 

Solution: 

Given, equation of hyperbola is 

x23y2=2x+8x22x3y2=8(x1)23y2=9(x1)29y23=1{{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 (x1)29+y23=1-\frac{{{(x-1)}^{2}}}{9}+\frac{{{y}^{2}}}{3}=1 and its eccentricity (e)=(a2+b2b2)(e)=\sqrt{\left( \frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} \right)}

Here,

a2=9b2=3e=9+33=2.{{a}^{2}}=9\\ {{b}^{2}}=3\\ e=\sqrt{\frac{9+3}{3}}=2.

Example 8: If the foci of the ellipse x216+y2b2=1\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 and the hyperbola x2144y281=125\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25} coincide, then find the value of b2{{b}^{2}}.

Solution:

Hyperbola is 

x2144y281=125a=14425,b=8125,e1=1+81144=225144=1512=54\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 =(ae1,0)=(125.54,0)=(3,0)=(a{{e}_{1}},0)=\left( \frac{12}{5}.\frac{5}{4},0 \right)=(3,\,0)

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

Hence b2=16(1916)=7.{{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)(1e)2+(1e)2=1B)1e+1e=1C)(1e)2+(1e)2=0D)1e+1e=2A) {{\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 x2a2y2b2=1(i)\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1 \rightarrow (i)

Then its conjugate will be, x2a2y2b2=1(ii)\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1 \rightarrow (ii)

If e is eccentricity of hyperbola (i), then

b2=a2(e21)1e2=a2(a2+b2)(iii){{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  

a2=b2(e21)1e2=b2(a2+b2)(iv){{a}^{2}}={{b}^{2}}(e{{‘}^{2}}-1)\\ \frac{1}{e{{‘}^{2}}}=\frac{{{b}^{2}}}{({{a}^{2}}+{{b}^{2}})} \rightarrow (iv)

Adding (iii) and (iv),

1(e)2+1e2=a2a2+b2+b2a2+b2=1.\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 25x216y2=400,25{{x}^{2}}-16{{y}^{2}}=400, whose mid point is (5, 3)?

Solution:

According to question, S25x216y2400=0S\equiv \,25{{x}^{2}}-16{{y}^{2}}-400=0

Equation of required chord is S1=T(i){{S}_{1}}=T \rightarrow (i)

Here,

S1=25(5)216(3)2400=625144400=81{{S}_{1}}=25{{(5)}^{2}}-16{{(3)}^{2}}-400 \\ =625-144-400=81 and 

T25xx116yy1400,T\equiv 25x{{x}_{1}}-16y{{y}_{1}}-400, where 

x1=5,y1=3=25(x)(5)16(y)(3)400=125x48y400{{x}_{1}}=5,\,{{y}_{1}}=3\\ =25(x)(5)-16(y)(3)-400\\ =125x-48y-400

So from (i), required chord is 

125x48y400=81125x48y=481.125x-48y-400=81\\ 125x-48y=481.