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.

---

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: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$

---

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.

---

• 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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