Chapter 11.13 - Solved Examples on Hyperbola

In the previous section, we saw latus rectum of hyperbola. We also saw a solved example. In this section, we will see a few more solved examples.

Solved example 11.15
Find the equation of the hyperbola with foci (0,±3) and vertices $\left(0,\pm \frac{\sqrt{11}}{2} \right)$
Solution:
1. The given foci are: (0,3) and (0,-3)
   ♦ These foci lie on the y-axis.
• So the equation of the hyperbola is of the form: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
2. Since the foci are (0,3) and (0,-3), we get: c = 3
3. The given vertices are: $\left(0, \frac{\sqrt{11}}{2} \right),~\left(0, -\frac{\sqrt{11}}{2} \right)$
• So we get: a = $\frac{\sqrt{11}}{2}$
4. Now we have 'a' and 'c'. We can calculate 'b'.
• We have: c² = a² + b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{3^2}
& {~=~}& {\left(\frac{\sqrt{11}}{2} \right)^2~+~b^2}
&{} \\

{\Rightarrow}&{9}
& {~=~}& {\frac{11}{4}~+~b^2}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {9~-~\frac{11}{4}}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {\frac{25}{4}}
&{} \\

{\Rightarrow}&{b}
& {~=~}& {\frac{5}{2}}
&{} \\

\end{array}$

• So the value of b is $\frac{5}{2}$ units.

5. Now we have 'a' and 'b'.
• Based on step (1), we can write:
Equation of the hyperbola is: $\frac{y^2}{\left(\frac{\sqrt{11}}{2} \right)^2}~-~\frac{x^2}{\left(\frac{5}{2} \right)^2}~=~1$
• This can be simplified as follows:

$\begin{array}{ll}
{}&{\frac{y^2}{\left(\frac{\sqrt{11}}{2} \right)^2}~-~\frac{x^2}{\left(\frac{5}{2} \right)^2}}
& {~=~}& {1}
&{} \\

{\Rightarrow}&{\frac{y^2}{\frac{11}{4}}~-~\frac{x^2}{\frac{25}{4}}}
& {~=~}& {1}
&{} \\

{\Rightarrow}&{\frac{4 y^2}{11}~-~\frac{4 x^2}{25}}
& {~=~}& {1}
&{} \\

{\Rightarrow}&{100 y^2~-~44 x^2}
& {~=~}& {275}
&{} \\

\end{array}$

Solved example 11.16
Find the equation of the hyperbola with foci (0,±12) and length of latus rectum 36.
Solution:
1. The given foci are: (0,12) and (0,-12)
   ♦ These foci lie on the y-axis.
• So the equation of the hyperbola is of the form: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
2. Since the foci are (0,12) and (0,-12), we get: c = 12
3. We have: Length of latus rectum = $\frac{2 b^2}{a}$ = 36
• From this, we get:
$\begin{array}{ll}
{}&{36}
& {~=~}& {\frac{2 × b^2}{a}}
&{} \\

{\Rightarrow}&{18}
& {~=~}& {\frac{b^2}{a}}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {18a}
&{} \\

\end{array}$

4. Now we have 'b²' in terms of 'a' . We can calculate 'a'.
• We have: c² = a² + b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{12^2}
& {~=~}& {a^2~+~18 a}
&{} \\

{\Rightarrow}&{144}
& {~=~}& {a^2~+~18a}
&{} \\

{\Rightarrow}&{a^2~+~18a~-~144}
& {~=~}& {0}
&{} \\

\end{array}$

• Solving this quadratic equation, we get: a = 6 or a = -24.
'a' is a length. It cannot be -ve. So we can write: a = 6

5. From the result in (3), we get: b² = 18a = 18 × 6 = 108

6. Now we have 'a' and 'b²'.
• Based on step (1), we can write:
Equation of the hyperbola is: $\frac{y^2}{36}~-~\frac{x^2}{108}~=~1$

Solved example 11.17
If the foci of the ellipse $\frac{x^2}{25}~+~\frac{y^2}{k^2}~=~1$ and the hyperbola $\frac{x^2}{144}~-~\frac{y^2}{81}~=~\frac{1}{25}$ coincide, find the value of k.
Solution:
1. We are given the complete equation of the hyperbola. So we will analyze it first.
• Given equation of the hyperbola is: $\frac{x^2}{144}~-~\frac{y^2}{81}~=~\frac{1}{25}$.
• It can be rearranged as follows:
$\begin{array}{ll}
{}&{\frac{x^2}{144}~-~\frac{y^2}{81}}
& {~=~}& {\frac{1}{25}}
&{} \\

{\Rightarrow}&{\frac{25 x^2}{144}~-~\frac{25 y^2}{81}}
& {~=~}& {\frac{25}{25}}
&{} \\

{\Rightarrow}&{\frac{x^2}{144/25}~-~\frac{y^2}{81/25}}
& {~=~}& {1}
&{} \\

{\Rightarrow}&{\frac{x^2}{(12/5)^2}~-~\frac{y^2}{(9/5)^2}}
& {~=~}& {1}
&{} \\

\end{array}$

2. In the above result, x² is the +ve term. So we can write:
The given hyperbola is of the form: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$ 
• Thus we get: a = 12/5 and b = 9/5
3. For any hyperbola, we have: c² = a² + b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{c^2}
& {~=~}& {\left(\frac{12}{5} \right)^2~+~\left(\frac{9}{5} \right)^2}
&{} \\

{}&{}
& {~=~}& {\frac{144}{25}~+~\frac{81}{25}}
&{} \\

{}&{}
& {~=~}& {\frac{225}{25}}
&{} \\

\end{array}$

• So the value of c is 15/5 = 3

4. The foci are (-c,0) and (c,0)
• So we get: (-3,0) and (3,0)

5. Given that, foci of the ellipse and the hyperbola are the same. So we can write:
• Foci of the ellipse $\frac{x^2}{25}~+~\frac{y^2}{k^2}~=~1$ are: (-3,0) and (3,0)
• Thus we can write: Value of 'c' for the ellipse is 3.   
6. Given equation of the ellipse is: $\frac{x^2}{25}~+~\frac{y^2}{k^2}~=~1$
• This is of the form: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
• So we can write: a = 5 and b = k
7. For any ellipse, we have: c² = a² - b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{3^2}
& {~=~}& {5^2~-~k^2}
&{} \\

{\Rightarrow}&{k^2}
& {~=~}& {5^2~-~3^2}
&{} \\

{\Rightarrow}&{k^2}
& {~=~}& {25~-~9}
&{} \\

{\Rightarrow}&{k^2}
& {~=~}& {16}
&{} \\

{\Rightarrow}&{k}
& {~=~}& {4}
&{} \\

\end{array}$

8. So we can write:
   ♦ The ellipse $\frac{x^2}{25}~+~\frac{y^2}{4^2}~=~1$
   ♦ And the hyperbola $\frac{x^2}{144}~-~\frac{y^2}{81}~=~\frac{1}{25}$
   ♦ Have the same foci (-3,0) and (3,0).
9. The actual plot is shown in fig.11.54 below:

Fig.11.54

 

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Link to a few more solved examples is given below:

Exercise 11.4

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In the next section, we will see some miscellaneous examples.

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Chapter 11.12 - Latus Rectum of Hyperbola

In the previous section, we completed a discussion on the standard equations of hyperbola. In this section, we will see latus rectum of hyperbola.

Latus rectum of a hyperbola

This can be explained in three steps:
1. Latus rectum is a line segment.
2. If a line segment is to qualify as the latus rectum of a hyperbola, it must satisfy three conditions.
(i) It must pass through F₁ or F₂.
(ii) It must be perpendicular to the transverse axis.
(iii) It’s end points must lie on the hyperbola.
3. Line segments AB and CD in fig.11.52 below satisfy all three conditions. So both are latus rectum of that hyperbola.

Fig.11.52

 

---

Length of the latus rectum

• We have seen the two forms where the equation of hyperbola is the simplest. Length of the latus rectum can be calculated very easily for those two forms.

• Let us see Form 1. It is shown in fig.11.52 above. We want the length AB. It can be calculated in 2 steps:
1. In the fig.11.52 above, let the length AF₂ be l.
• Then the coordinates of A will be (c,l)
2. Point A lies on the hyperbola. So we can write:

$\begin{array}{ll}
{}&{\frac{c^2}{a^2}~-~\frac{l^2}{b^2}}
& {~=~}& {1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{c^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{a^2~+~b^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{a^2}{a^2}~+~\frac{b^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {1~+~\frac{b^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{b^2}{a^2}}
&{} \\

{\Rightarrow}&{l^2}
& {~=~}& {\frac{b^4}{a^2}}
&{} \\

{\Rightarrow}&{l}
& {~=~}& {\frac{b^2}{a}}
&{} \\

{\Rightarrow}&{2l}
& {~=~}& {\frac{2b^2}{a}}
&{} \\

\end{array}$

• Note:
In the above calculation, first we obtained l. Then we doubled it to obtain the total length AB. This is because, the hyperbola is symmetric about the transverse axis.

◼ So we can write:
Length of the latus rectum of the hyperbola $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$ is  $\frac{2 b^2}{a}$

---

• Let us see Form 2. It is shown in fig.11.53 below:

Fig.11.53

• We want the length AB. It can be calculated in 2 steps:
1. In the fig.11.53 above, let the length AF₁ be l.
Then the coordinates of A will be (-l,c)
2. Point A lies on the hyperbola. So we can write:

$\begin{array}{ll}
{}&{\frac{c^2}{a^2}~-~\frac{(-l)^2}{b^2}}
& {~=~}& {1}
&{} \\

{}&{\frac{c^2}{a^2}~-~\frac{l^2}{b^2}}
& {~=~}& {1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{c^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{a^2~+~b^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {1~+~\frac{b^2}{a^2}~-~1}
&{} \\

{\Rightarrow}&{l^2}
& {~=~}& {\frac{b^4}{a^2}}
&{} \\

{\Rightarrow}&{l}
& {~=~}& {\frac{b^2}{a}}
&{} \\

{\Rightarrow}&{2l}
& {~=~}& {\frac{2b^2}{a}}
&{} \\

\end{array}$

• Note:
In the above calculation, first we obtained l. Then we doubled it to obtain the total length AB. This is because, the hyperbola is symmetric about the transverse axis.

◼ So we can write:
Length of the latus rectum of the hyperbola $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$ is also  $\frac{2 b^2}{a}$

---

Now we will see a solved example:

Solved example 11.14
For the hyperbolas:
(a) $\frac{x^2}{9}~-~\frac{y^2}{16}~=~1$
(b) y² - 16x² = 16
find the following:
(i) coordinates of the foci
(ii) coordinates of the vertices
(iii) the eccentricity
(iv) length of the latus rectum.
Solution:
Part (a):
1. x² has the +ve term.
• So the equation of the hyperbola is of the form: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
• So we can write:
    ♦ Transverse axis of this hyperbola lies along the x-axis.
    ♦ Conjugate axis of this hyperbola lies along the y-axis.
    ♦ a² = 9. So a = 3
    ♦ b² = 16. So b = 4
2. We have: c² = a² + b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{c^2}
& {~=~}& {3^2~+~4^2}
&{} \\

{}&{}
& {~=~}& {9~+~16}
&{} \\

{}&{}
& {~=~}& {25}
&{} \\

\end{array}$

• So the value of c is 5

3. The coordinates of the foci are (-c,0) and (c,0)
• So in our present case, the coordinates are: (-5,0) and (5,0)
• This is the answer for part (i).
4. The coordinates of the vertices are (-a,0) and (a,0)
• So in our present case, the coordinates are: (-3,0) and (3,0)
• This is the answer for part (ii).
5. Eccentricity is given by: e = c/a
• So in our present case, e = 5/3
• This is the answer for part (iii).
6. We have: Length of latus rectum = $\frac{2 b^2}{a}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\text{Length}}
& {~=~}& {\frac{2 × 4^2}{3}}
&{} \\

{}&{}
& {~=~}& {\frac{32}{3}~\text{units}}
&{} \\

\end{array}$
• This is the answer for part (iv).

Part (b):
• The given equation is y² - 16x² = 16.
   ♦ Numerator of the coefficient of x² must be 1.
   ♦ Numerator of the coefficient of y² must be 1.
   ♦ Right side of the equation must be 1.
• So we divide the given equation by 16. We get: $\frac{y^2}{16}~-~\frac{x^2}{1}~=~1$
1. y² has the +ve term.
• So the equation of the hyperbola is of the form: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
• So we can write:
    ♦ Transverse axis of this hyperbola lies along the y-axis.
    ♦ Conjugate axis of this hyperbola lies along the x-axis.
    ♦ a² = 16. So a = 4
    ♦ b² = 1. So b = 1
2. We have: c² = a² + b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{c^2}
& {~=~}& {4^2~+~^2}
&{} \\

{}&{}
& {~=~}& {16~+~1}
&{} \\

{}&{}
& {~=~}& {17}
&{} \\

\end{array}$

• So the value of c is √17

3. The coordinates of the foci are (0,-c) and (0,c)
• So in our present case, the coordinates are: (0,-√17) and (0,√17)
• This is the answer for part (i).
4. The coordinates of the vertices are (-a,0) and (a,0)
• So in our present case, the coordinates are: (0,-4) and (0,4)
• This is the answer for part (ii).
5. Eccentricity is given by: e = c/a
• So in our present case, e = (√17)/4
• This is the answer for part (iii).
6. We have: Length of latus rectum = $\frac{2 b^2}{a}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\text{Length}}
& {~=~}& {\frac{2 × 1^2}{4}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2}~\text{units}}
&{} \\

\end{array}$
• This is the answer for part (iv).

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In the next section, we will see a few more solved examples.

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Chapter 11.10 - Simplest Equation of A Hyperbola

In the previous section, we saw the basic properties of a hyperbola. In this section, we will see equation of an hyperbola.

• To write the equation of a hyperbola, we must first place it on the Cartesian plane.
• The equation will be in the simplest form when the following three conditions are satisfied:
    ♦ The center of the hyperbola is at the origin O.
    ♦ The transverse axis of the hyperbola lies along the x-axis.
    ♦ The conjugate axis of the hyperbola lies along the y-axis.
• This is shown in fig.11.45 below:

Fig.11.45

• Based on fig.11.45, we can derive the equation in 6 steps:
1. Let P(x,y) be any point on the hyperbola.
2. We know that, F₁ is at a distance of ‘c’ from O.
• So the coordinates of F₁ will be (-c,0)   
3. We know that, F₂ is at a distance of ‘c’ from O.
• So the coordinates of F₂ will be (c,0)
4. Now we have three points and their coordinates:
P(x,y), F₁(-c,0), F₂(c,0)
• Using the distance formula, we can write some distances:

• First we write the distance PF₁:
$\begin{array}{ll}
{}&{PF_1}
& {~=~}& {\sqrt{(x~-~ -c)^2~+~(y - 0)^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x + c)^2~+~y^2}}
&{} \\

\end{array}$ 

• Next we write the distance PF₂:
$\begin{array}{ll}
{}&{PF_2}
& {~=~}& {\sqrt{(x~-~ c)^2~+~(y - 0)^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x - c)^2~+~y^2}}
&{} \\

\end{array}$

• Difference of the above two distances is: $\sqrt{(x + c)^2~+~y^2}~-~\sqrt{(x - c)^2~+~y^2}$

5. We know that, the constant difference for a hyperbola is '2a'

6. Equating the results in (4) and (5), we get:

$\begin{array}{ll}
{}&{\sqrt{(x + c)^2~+~y^2}~-~\sqrt{(x - c)^2~+~y^2}}
& {~=~}& {2a}
&{} \\

{\Rightarrow}&{\sqrt{(x + c)^2~+~y^2}}
& {~=~}& {2a~+~\sqrt{(x - c)^2~+~y^2}}
&{} \\

{\Rightarrow}&{(x + c)^2~+~y^2}
& {~=~}& {4a^2~+~4a \sqrt{(x - c)^2~+~y^2}~+~(x - c)^2~+~y^2~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{x^2 + 2xc + c^2 + y^2}
& {~=~}& {4a^2~+~4a \sqrt{(x - c)^2~+~y^2}~+~x^2 - 2xc + c^2~+~y^2}
&{} \\

{\Rightarrow}&{2xc}
& {~=~}& {4a^2~+~4a \sqrt{(x - c)^2~+~y^2}- 2xc}
&{} \\

{\Rightarrow}&{4xc}
& {~=~}& {4a^2~+~4a \sqrt{(x - c)^2~+~y^2}}
&{} \\

{\Rightarrow}&{xc}
& {~=~}& {a^2~+~a \sqrt{(x - c)^2~+~y^2}~~ \color {green} {\text{- - - (II)}}}
&{} \\

{\Rightarrow}&{\frac{xc}{a}}
& {~=~}& {a~+~\sqrt{(x - c)^2~+~y^2}}
&{} \\

{\Rightarrow}&{\sqrt{(x - c)^2~+~y^2}}
& {~=~}& {\frac{xc}{a}~-~a~~ \color {green} {\text{- - - (III)}}}
&{} \\

{\Rightarrow}&{(x - c)^2~+~y^2}
& {~=~}& {\frac{x^2 c^2}{a^2}~-~\frac{2axc}{a}~+~a^2}
&{} \\

{\Rightarrow}&{(x - c)^2~+~y^2}
& {~=~}& {a^2~-~2cx~+~\frac{x^2 c^2}{a^2}}
&{} \\

{\Rightarrow}&{x^2 - 2cx + c^2~+~y^2}
& {~=~}& {a^2~-~2cx~+~\frac{x^2 c^2}{a^2}}
&{} \\

{\Rightarrow}&{x^2 + c^2~+~y^2}
& {~=~}& {a^2~+~\frac{x^2 c^2}{a^2}}
&{} \\

{\Rightarrow}&{x^2 ~+~y^2~-~\frac{x^2 c^2}{a^2}}
& {~=~}& {a^2 - c^2}
&{} \\

{\Rightarrow}&{x^2 \left(1~-~\frac{c^2}{a^2} \right)~+~y^2}
& {~=~}& {a^2 - c^2}
&{} \\

{\Rightarrow}&{x^2 \left(\frac{a^2~-~c^2}{a^2} \right)~+~y^2}
& {~=~}& {a^2 - c^2~~ \color {green} {\text{- - - (IV)}}}
&{} \\

{\Rightarrow}&{x^2 \left(\frac{c^2~-~a^2}{a^2} \right)~-~y^2}
& {~=~}& {c^2 - a^2~~ \color {green} {\text{- - - (V)}}}
&{} \\

{\Rightarrow}&{x^2 \left(\frac{b^2}{a^2} \right)~-~y^2}
& {~=~}& {b^2~~ \color {green} {\text{- - - (VI)}}}
&{} \\

{\Rightarrow}&{\frac{x^2}{a^2}~-~\frac{y^2}{b^2}}
& {~=~}& {1}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as (I):
In this line, we square both sides.
• Line marked as (II):
In this line, we divide both sides by 4.
• Line marked as (III):
In this line, we square both sides.
• Line marked as (IV):
In this line, multiply both sides by -1.
• Line marked as (V):
In this line, write b² in the place of c² - a².
• Line marked as (VI):
In this line, we divide both sides by b².

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Using the above 6 steps, we derived an equation. Now we will prove the converse. It can be written in 11 steps:

1. We derived an equation: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
2. To prove the converse, we assume a point P.
• Let P(x,y) be any point on the ellipse.
• Distance of P from F₁ can be written as:

$\begin{array}{ll}
{}&{PF_1}
& {~=~}& {\sqrt{(x~-~ -c)^2~+~(y - 0)^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x + c)^2~+~y^2}}
&{} \\

\end{array}$

3. But based on the equation written in (1), we can write:

$\begin{array}{ll}
{}&{\frac{y^2}{b^2}}
& {~=~}& {\frac{x^2}{a^2}-1}
&{} \\

{\Rightarrow}&{y^2}
& {~=~}& {b^2\left(\frac{x^2}{a^2}~-~1 \right)}
&{} \\

\end{array}$

4. Substituting the above result in (2), we get:

$\begin{array}{ll}
{}&{PF_1}
& {~=~}& {\sqrt{(x + c)^2~+~y^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x + c)^2~+~b^2\left(\frac{x^2}{a^2}~-~1 \right)}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x + c)^2~+~\left(c^2 - a^2 \right) \left(\frac{x^2 }{a^2}~-~1 \right)}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {\sqrt{x^2 + 2cx + c^2~+~\frac{c^2 x^2}{a^2} - c^2 - x^2 + a^2 }}
&{} \\

{}&{}
& {~=~}& {\sqrt{2cx~+~a^2 + \frac{c^2 x^2}{a^2}}}
&{} \\

{}&{}
& {~=~}& {\sqrt{\left(a+\frac{cx}{a} \right)^2}}
&{} \\

{}&{}
& {~=~}& {a+\frac{cx}{a}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as (I):
In this line, we write c² - a² in the place of b².

5. Now we consider the distance of P from F₂. It can be written as:

$\begin{array}{ll}
{}&{PF_2}
& {~=~}& {\sqrt{(x~-~ c)^2~+~(y - 0)^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x - c)^2~+~y^2}}
&{} \\

\end{array}$

6. As we did in the case of PF₁, here also, we substitute for y². We get:

$\begin{array}{ll}
{}&{PF_2}
& {~=~}& {\sqrt{(x - c)^2~+~y^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x - c)^2~+~b^2\left(\frac{x^2 - a^2}{a^2} \right)}}
&{} \\

{}&{}
& {~=~}& {\sqrt{(x - c)^2~+~\left(c^2 - a^2 \right) \left(\frac{x^2 }{a^2}~-~1 \right)}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {\sqrt{x^2 - 2cx + c^2~+~\frac{c^2 x^2}{a^2} - c^2 - x^2 + a^2 }}
&{} \\

{}&{}
& {~=~}& {\sqrt{-2cx~+~a^2 + \frac{c^2 x^2}{a^2}}}
&{} \\

{}&{}
& {~=~}& {\sqrt{\left(a-\frac{cx}{a} \right)^2}}
&{} \\

{}&{}
& {~=~}& {a-\frac{cx}{a}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as (I):
In this line, we write c² - a² in the place of b². 

7. Now we have to find the difference between PF₁ and PF₂. It can be calculated in 6 steps:
(i) We know that, the vertices are at a distance of 'a' from O.
• So we can draw the vertical lines x= -a and x = a through the vertices. This is shown in fig.11.46 below:

Fig.11.46

(ii) Our point P is on the right branch of the hyperbola.
• That means, P is on the right side of the line x=a
    ♦ So the x-coordinate of P will be greater than 'a'.
    ♦ We can write: x > a
(iii) Now, $\frac{c}{a}$ is greater than '1' because, 'c' is greater than 'a'.
    ♦ Since $\frac{c}{a}$ is greater than '1', and x>a, we can write: $\frac{cx}{a}$ is greater than 'a'.
(iv) So the distance PF₂ = $a-\frac{cx}{a}$ will become -ve
• To make it +ve, we must write: PF₂ = $\frac{cx}{a}-a$
(v) The distance PF₁ = $a+\frac{cx}{a}$ need not be adjusted because, subtraction is not involved. 
(vi) Now we can write the difference:
PF₁ - PF₂ = $a+\frac{cx}{a}~-~\left( \frac{cx}{a}-a \right)$
= $a+\frac{cx}{a} -  \frac{cx}{a}+ a$ = 2a
8. In the step (7) above, we considered the case when the point P is on the right side branch of the hyperbola.
• Fig.11.47 below shows the case when P is on the left side branch.

Fig.11.47

• Here also, to find the difference, we must make some adjustments. It can be written in 6 steps:
(i) We know that, the vertices are at a distance of 'a' from O.
• So we can draw the vertical lines x= -a and x = a through the vertices. This is shown in fig.11.47 above.
(ii) Our point P is on the left branch of the hyperbola.
• That means, P is on the left side of the line x=-a
    ♦ So the x-coordinate of P will be lesser than '-a'.
    ♦ We can write: x will be -ve
(iii) Since x is -ve, the distance PF₂ = $a-\frac{cx}{a}$ will become +ve. So there is no adjustment required for PF₂
(iv) Now consider PF₁ = $a+\frac{cx}{a}$
• c/a is greater than '1' because, 'c' is always greater than 'a'.
    ♦ Since $\frac{c}{a}$ is greater than '1', and 'x' is -ve, we can write: $\frac{cx}{a}$ is -ve and numerically greater than 'a'.
(v) So the distance PF₁ = $a+\frac{cx}{a}$ will become -ve
• To make it +ve, we must write: PF₁ = $-\left(a+\frac{cx}{a} \right)$
(vi) Now we can write the difference:
• Since PF₂ is larger, we must subtract PF₁ from PF₂.
PF₂ - PF₁ = $\left(a-\frac{cx}{a}\right)~-~-\left(a+\frac{cx}{a} \right)$
= $a - \frac{cx}{a}+ a + \frac{cx}{a}$ = 2a
9. Based on steps (7) and (8), we can write:
The point P can be anywhere on the hyperbola. The difference will always be '2a'.
10. In the previous section, we saw that, the constant difference of the hyperbola is '2a'.
11. So any point P(x,y) on the hyperbola will satisfy the equation $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
• The converse is proved.

◼ So we can write:
If the center of the hyperbola is at O, transverse axis lies along the x-axis and conjugate axis lies along the y-axis, then equation of the hyperbola is: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$

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Based on the above equation of the hyperbola, we can write an interesting fact. It can be written in 4 steps:
1. We have: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
2. This can be rearranged as: $\frac{x^2}{a^2}~=~1~+~\frac{y^2}{b^2}$
• So $\frac{x^2}{a^2}$ will be always greater than 1.
• That is: $\left|\frac{x}{a}\right|~\ge~1$
3. Solving the above inequality, we get:
    ♦ x should not be greater than -a.
    ♦ x should not be less than a.
• That is: $x \le -a ~\text{or}~x \ge a$
4. So we can write:
• Consider any point on the hyperbola. It will not lie between the two vertical lines:
    ♦ x = -a.
    ♦ x =  a.

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• In this section, we saw a simplest equation of a hyperbola.
    ♦ The center is at O.
    ♦ The transverse axis lies along the x-axis.
    ♦ The conjugate axis lies along the y-axis.
• We will get another simplest equation also when:
    ♦ The center is at O.
    ♦ The transverse axis lies along the y-axis.
    ♦ The conjugate axis lies along the x-axis.
• We will see it in the next section.

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Chapter 11.9 - Hyperbola

In the previous section, we completed a discussion on ellipse. In this section, we will see hyperbola.

Some basics about hyperbola can be written in 6 steps:
1. Consider the five points F₁, F₂, P₁, P₂ and P₃ marked in fig.11.42 below:

Fig.11.42

2. Let us write some distances:
• Distance of P₁:
   ♦ Distance of P₁ from F₁ is 8.6 units.
   ♦ Distance of P₁ from F₂ is 18.7 units.
         ✰ The difference of the two distances is (18.7 - 8.6) = 10.1 units.
• Distance of P₂:
   ♦ Distance of P₂ from F₁ is 16.7 units.
   ♦ Distance of P₂ from F₂ is 6.6 units.
         ✰ The difference of the two distances is (16.7 - 6.6) = 10.1 units.
• Distance of P₃:
   ♦ Distance of P₃ from F₁ is 14.1 units.
   ♦ Distance of P₃ from F₂ is 4.0 units.
         ✰ The difference of the two distances is (14.1 - 4.0) = 10.1 units.
3. So the three points P₁, P₂ and P₃ have a specialty. It can be written in two steps:
(i) Take any one of those three points.
   ♦ Measure the distance of that point from F₁.
   ♦ Measure the distance of that point from F₂.
(ii) The difference of the two distances will be 10.1 units.
4. There are infinite number of points for which the difference is 10.1 units.      
• All such points will lie in the red curve.
• The red curve is called a hyperbola.
5. So we can write the definition:
A hyperbola is the set of all points in a plane difference of whose distances from two fixed points in the plane is a constant.
6. Note that, for a particular hyperbola, the two points F₁ and F₂ are fixed. If we change one or both of those points, we will get another hyperbola.

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Now we will see some basic features of hyperbola. They can be written in 8 steps:
1. We have seen that, F₁ and F₂ are two fixed points.
• They are called the foci of the hyperbola.
(‘foci’ is the plural of ‘focus’)
2. Draw a line connecting the two foci.
• Let it intersect the hyperbola at A and B.
• Then the line through the foci is called the transverse axis of the hyperbola.
• This is shown in fig.11.43 below:

Fig.11.43

3. The midpoint of F₁F₂ is called center of the hyperbola. It is denoted by the letter O.
4. Draw a line through O and perpendicular to AB.
• This line is called the conjugate axis of the hyperbola.
5. The points A and B at which the transverse axis meets the hyperbola are called vertices of the hyperbola.
6. The length AB is written as ‘2a’. This is shown in fig.11.44 below:

Fig.11.44

• The distance between the two foci is written as ‘2c’
7. Based on the above fig.11.44, we can write:
   ♦ Distance from center to any one vertex is 'a'.
   ♦ '2a' is the length of the transverse axis.
   ♦ Distance from center to any one focus is 'c'.
8. '2b' is considered as the length of the conjugate axis.
   ♦ Where $b=\sqrt{c^2 - a^2}$

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In fig.11.42, we saw that, the difference of the distances is a constant. Our next aim is to find the value of that constant. It can be written in 3 steps:
1. Consider a point P on the hyperbola.
• Let it be situated at the vertex A
• Then the distance of P from F₁ = AF₁ = (OF₁ – OA) = c-a
• Also the distance of P from F₂ = AF₂ = (OA + OB + BF₂)
= [a+a+(c-a)] = c + a  
2. So the difference between the distances = [(c+a) – (c-a)] = [c+a-c+a] = 2a
3. We can write:
For any point P on the hyperbola, the difference in distances from the foci will be ‘2a’          

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Eccentricity of a Hyperbola

This can be explained in 5 steps:
1. Consider the distance ‘c’.
• It is the distance of the focus from the center O.
2. Consider the distance ‘a’.
• It is the distance of the vertex from the center O.
3. Eccentricity of a hyperbola is the ratio of the above two items. It is denoted by the letter ‘e’.
So we can write: $\rm{e=\frac{c}{a}}$
4. Based on this result, we can write: c = ae.
• That means, in any hyperbola, the focus is at a distance of ae from the center.
5. From fig.11.44, it is clear that, 'c' is greater than 'a'. So the eccentricity is never less than 1.

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In the next section, we will see the standard equations of a hyperbola.

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