In the previous section,
we saw the simplest equation of a hyperbola.
♦ The center was at O.
♦ The transverse axis was along the x-axis.
♦ The conjugate axis was along the y-axis.
•
To write the equation of a hyperbola, we must first place it on the
Cartesian plane.
• The equation will be in the simplest form also when the
following three conditions are satisfied:
♦ The center is at O.
♦ The transverse axis lies along the y-axis.
♦ The conjugate axis lies along the x-axis.
• This is shown in fig.11.48 below:
 |
| Fig.11.48 |
• Based on fig.11.48, 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 (0,c)
3. We know that, F2 is at a distance of ‘c’ from O.
• So the coordinates of F2 will be (0,-c)
4. Now we have three points and their coordinates:
P(x,y), F1(0,c), F2(0,-c)
• Using the distance formula, we can write some distances:
• First we write the distance PF1:
$\begin{array}{ll}
{}&{PF_1}
& {~=~}& {\sqrt{(x~-~ 0)^2~+~(y - c)^2}}
&{} \\
{}&{}
& {~=~}& {\sqrt{x^2~+~(y-c)^2}}
&{} \\
\end{array}$
• Next we write the distance PF2:
$\begin{array}{ll}
{}&{PF_2}
& {~=~}& {\sqrt{(x~-~ 0)^2~+~(y~-~-c)^2}}
&{} \\
{}&{}
& {~=~}& {\sqrt{x^2~+~(y+c)^2}}
&{} \\
\end{array}$
• Difference of the above two distances is: $\sqrt{x^2~+~(y-c)^2}~-~\sqrt{x^2~+~(y+c)^2}$
5. We know that, the constant difference for a hyperbola is '2a'
6. Equating the results in (4) and (5), we get:
$\begin{array}{ll}
{}&{\sqrt{x^2~+~(y-c)^2}~-~\sqrt{x^2~+~(y+c)^2}}
& {~=~}& {2a}
&{} \\
{\Rightarrow}&{\sqrt{x^2~+~(y-c)^2}}
& {~=~}& {2a~+~\sqrt{x^2~+~(y+c)^2}}
&{} \\
{\Rightarrow}&{x^2~+~(y-c)^2}
& {~=~}& {4a^2~+~4a \sqrt{x^2~+~(y+c)^2}~+~x^2~+~(y+c)^2~~ \color {green} {\text{- - - (I)}}}
&{} \\
{\Rightarrow}&{x^2 + y^2 - 2yc + c^2}
& {~=~}& {4a^2~+~4a \sqrt{x^2~+~(y+c)^2}~+~x^2 + y^2 + 2yc + c^2}
&{} \\
{\Rightarrow}&{-2yc}
& {~=~}& {4a^2~+~4a \sqrt{x^2~+~(y+c)^2} + 2yc}
&{} \\
{\Rightarrow}&{-4yc}
& {~=~}& {4a^2~+~4a \sqrt{x^2~+~(y+c)^2}}
&{} \\
{\Rightarrow}&{-yc}
& {~=~}& {a^2~+~a \sqrt{x^2~+~(y+c)^2}~~ \color {green} {\text{- - - (II)}}}
&{} \\
{\Rightarrow}&{-\frac{yc}{a}}
& {~=~}& {a~+~\sqrt{x^2~+~(y+c)^2}}
&{} \\
{\Rightarrow}&{\sqrt{x^2~+~(y+c)^2}}
& {~=~}& {a~+~\frac{yc}{a}~~ \color {green} {\text{- - - (III)}}}
&{} \\
{\Rightarrow}&{x^2~+~(y+c)^2}
& {~=~}& {a^2~+~\frac{2ayc}{a}~+~\frac{y^2 c^2}{a^2}}
&{} \\
{\Rightarrow}&{x^2~+~(y+c)^2}
& {~=~}& {a^2~+~2yc~+~\frac{y^2 c^2}{a^2}}
&{} \\
{\Rightarrow}&{x^2 + y^2 + 2yc + c^2}
& {~=~}& {a^2~+~2yc~+~\frac{y^2 c^2}{a^2}}
&{} \\
{\Rightarrow}&{x^2 + y^2 + c^2}
& {~=~}& {a^2 + \frac{y^2 c^2}{a^2}}
&{} \\
{\Rightarrow}&{x^2 ~+~y^2~-~\frac{y^2 c^2}{a^2}}
& {~=~}& {a^2 - c^2}
&{} \\
{\Rightarrow}&{x^2 ~+~y^2 \left(1~-~\frac{c^2}{a^2} \right)}
& {~=~}& {a^2 - c^2}
&{} \\
{\Rightarrow}&{x^2 ~+~y^2 \left(\frac{a^2~-~c^2}{a^2} \right)}
& {~=~}& {a^2 - c^2~~ \color {green} {\text{- - - (IV)}}}
&{} \\
{\Rightarrow}&{-x^2 ~+~y^2 \left(\frac{c^2~-~a^2}{a^2} \right)}
& {~=~}& {c^2 - a^2~~ \color {green} {\text{- - - (V)}}}
&{} \\
{\Rightarrow}&{-x^2 ~+~y^2 \left(\frac{b^2}{a^2} \right)}
& {~=~}& {b^2~~ \color {green} {\text{- - - (VI)}}}
&{} \\
{\Rightarrow}&{\frac{y^2}{a^2}~-~\frac{x^2}{b^2}}
& {~=~}& {1}
&{} \\
\end{array}$
◼ 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, we multiply both sides by -1.
• Line marked as (V):
In this line, we 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 8 steps:
1. We derived an equation: $\frac{y^2}{a^2} - \frac{x^2}{b^2}~=~1$
2. To prove the converse, we assume a point P.
• Let P(x,y) be any point on the hyperbola.
• Distance of P from F1 can be written as:
$\begin{array}{ll}
{}&{PF_1}
& {~=~}& {\sqrt{(x~-~ 0)^2~+~(y - c)^2}}
&{} \\
{}&{}
& {~=~}& {\sqrt{x^2~+~(y-c)^2}}
&{} \\
\end{array}$
3. But based on the equation written in (1), we can write:
$\begin{array}{ll}
{}&{\frac{x^2}{b^2}}
& {~=~}& {\frac{y^2}{a^2}-1}
&{} \\
{\Rightarrow}&{x^2}
& {~=~}& {b^2\left(\frac{y^2}{a^2}- 1 \right)}
&{} \\
\end{array}$
4. Substituting the above result in (2), we get:
$\begin{array}{ll}
{}&{PF_1}
& {~=~}& {\sqrt{x^2~+~(y-c)^2}}
&{} \\
{}&{}
& {~=~}& {\sqrt{b^2\left(\frac{y^2}{a^2}-1 \right)~+~(y-c)^2}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\left(c^2 - a^2 \right) \left(\frac{y^2}{a^2}-1 \right)~+~(y-c)^2}~~ \color {green} {\text{- - - (I)}}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\frac{c^2 y^2}{a^2} - c^2 - y^2 + a^2 ~+~y^2 - 2yc + c^2}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\frac{c^2 y^2}{a^2} + a^2 - 2yc }}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\left( a - \frac{cy}{a} \right)^2}}
&{} \\
{}&{}
& {~=~}& {a - \frac{cy}{a}}
&{} \\
\end{array}$
◼ 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:
$\begin{array}{ll}
{}&{PF_2}
& {~=~}& {\sqrt{(x~-~ 0)^2~+~(y~-~-c)^2}}
&{} \\
{}&{}
& {~=~}& {\sqrt{x^2~+~(y+c)^2}}
&{} \\
\end{array}$
6. As we did in the case of PF1, here also, we substitute for x2. We get:
$\begin{array}{ll}
{}&{PF_2}
& {~=~}& {\sqrt{x^2~+~(y+c)^2}}
&{} \\
{}&{}
& {~=~}& {\sqrt{b^2\left(\frac{y^2}{a^2}-1 \right)~+~(y+c)^2}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\left(c^2 - a^2 \right) \left(\frac{y^2}{a^2}-1 \right)~+~(y+c)^2}~~ \color {green} {\text{- - - (I)}}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{ \frac{c^2 y^2}{a^2} - c^2 - y^2 + a^2~+~y^2 + 2yc + c^2}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\frac{c^2 y^2}{a^2} + a^2 + 2yc}}
&{} \\
{}&{}
&
{~=~}& {\sqrt{\left( a + \frac{cy}{a} \right)^2}}
&{} \\
{}&{}
& {~=~}& {a + \frac{cy}{a}}
&{} \\
\end{array}$
◼ 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 horizontal lines y= a and y = -a through the vertices. This is shown in fig.11.49 below:
 |
| Fig.11.49 |
(ii) Our point P is on the top branch of the hyperbola.
• That means, P is above the line y=a
♦ So the y-coordinate of P will be greater than 'a'.
♦ We can write: y > a
(iii) Now, $\frac{c}{a}$ is greater than '1' because, 'c' is always greater than 'a'.
♦ Since $\frac{c}{a}$ is greater than '1', and y>a, we can write: $\frac{cy}{a}$ is greater than 'a'.
(iv) So the distance PF1 = $a-\frac{cy}{a}$ will become -ve
• To make it +ve, we must write: PF1 = $\frac{cy}{a}-a$
(v) The distance PF2 = $a+\frac{cy}{a}$ need not be adjusted because, subtraction is not involved.
(vi) Now we can write the difference:
• Since PF2 is larger, we must subtract PF1 from PF2.
PF2 - PF1 = $a+\frac{cy}{a}~-~\left( \frac{cy}{a}-a \right)$
= $a+\frac{cy}{a} - \frac{cy}{a}+ a$ = 2a
8. In the step (7) above, we considered the case when the point P is on the top branch of the hyperbola.
• Fig.11.50 below shows the case when P is on the bottom branch.
 |
| Fig.11.50 |
• 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 horizontal lines y= -a and y = a through the vertices. This is shown in fig.11.50 above.
(ii) Our point P is on the bottom branch of the hyperbola.
• That means, P is below the line y=-a
♦ So the y-coordinate of P will be lesser than '-a'.
♦ We can write: y will be -ve
(iii) Since y is -ve, the distance PF1 = $a-\frac{cy}{a}$ will become +ve. So there is no adjustment required for PF1
(iv) Now consider PF2 = $a+\frac{cy}{a}$
• $\frac{c}{a}$ is greater than '1' because, 'c' is always greater than 'a'.
♦ Since $\frac{c}{a}$ is greater than '1', and 'y' is -ve, we can
write: $\frac{cy}{a}$ is -ve and numerically greater than 'a'.
(v) So the distance PF2 = $a+\frac{cy}{a}$ will become -ve
• To make it +ve, we must write: PF2 = $-\left(a+\frac{cy}{a} \right)$
(vi) Now we can write the difference:
• Since PF1 is larger, we must subtract PF2 from PF1.
PF1 - PF2 = $\left(a-\frac{cy}{a}\right)~-~-\left(a+\frac{cy}{a} \right)$
= $a - \frac{cy}{a}+ a + \frac{cy}{a}$ = 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 $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
• The converse is proved.
◼ So we can write:
If
the center of the hyperbola is at O, transverse axis lies along the
y-axis and conjugate axis lies along the x-axis, then equation of the
hyperbola is:
$\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
Based on the above equation of the hyperbola, we can write an interesting fact. It can be written in 4 steps:
1. We have: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
2. This can be rearranged as: $\frac{y^2}{a^2}~=~1~+~\frac{x^2}{b^2}$
• So $\frac{y^2}{a^2}$ will be always greater than 1.
• That is: $\left|\frac{y}{a}\right|~\ge~1$
3. Solving the above inequality, we get:
♦ y should not be greater than -a.
♦ y should not be less than a.
• That is: $y \le -a ~\text{or}~y \ge a$
4. So we can write:
• Consider any point on the hyperbola. It will not lie between the two horizontal lines:
♦ y = -a.
♦ y = a.
• So we have seen the two simplest forms of the hyperbola. Let us write a comparison between the two forms:
A. Comparison based on orientation:
• This can be written in 5 steps:
1. In the first form,
♦ Transverse axis lies along the x-axis.
♦ Conjugate axis lies along the y-axis.
2. In the second form,
♦ Transverse axis lies along the y-axis.
♦ Conjugate axis lies along the x-axis.
3. Whatever be the orientation,
♦ Length of semi-transverse axis is denoted by the letter ‘a’.
♦ Length of semi-conjugate axis is denoted by the letter ‘b'.
4. Equations of the two orientations:
♦ For the first form, the equation is: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
♦ For the second form, the equation is: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
• These are known as the standard equations of the hyperbola.
5. A hyperbola in which 'a' and 'b' are same, is called an equilateral hyperbola.
B. Comparison of coefficients:
• This can be written in 2 steps:
1. For the first form, $\frac{1}{a^2}$ is the coefficient of x2
♦ Here x2 is the +ve term.
2. For the second form, $\frac{1}{a^2}$ is the coefficient of y2
♦ Here y2 is the +ve term.
◼ So in the given equation,
• If x2 is the +ve term, then the transverse axis of the hyperbola lies along the x-axis.
• If y2 is the +ve term, then the transverse axis of the hyperbola lies along the y-axis.
C. Comparison based on symmetry:
• This can be written in 3 steps:
1. Whatever be the orientation, the hyperbola will be symmetric about the transverse axis and conjugate axis.
2. This is because, the x and y values are being squared.
♦ +ve x and -ve x give the same result.
♦ +ve y and -ve y give the same result.
3. So four symmetric combinations are possible:
(x,y), (-x,y), (x,-y) and (-x,-y).
• An example is shown in fig.11.51 below:
 |
| Fig.11.51 |
So we have seen the standard equations of the hyperbola. In the next section, we will see latus rectum of hyperbola.
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