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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Chapter 11.2 - Simplest Equation of Parabola

In the previous section, we saw the details about circles. In this section, we will see the details about parabolas.

Some basics about parabola can be written in 9 steps:
1. Consider the point P₁ in fig.11.15 below:

Fig.11.15

• This point P₁ satisfies one condition. It can be written in three steps:
(i) The distance of P₁ from the magenta line is P₁P’₁
(Remember that, the distance of a point from a line should be measured in the perpendicular direction. That is., the line segment P₁P’₁ must be perpendicular to the magenta line.)
(ii) The distance of P₁ from the point F is FP₁
(iii) The condition is that, the above two distances are equal.
• That is., P₁P’₁ = FP₁
    ♦ In the fig., both distances are 11.8 units 
2. Consider the point P₂ in the fig.11.15.
This point P₂ satisfies one condition. It can be written in steps:
(i) The distance of P₂ from the magenta line is P₂P’₂
(ii) The distance of P₂ from the point F is FP₂
(iii) The condition is that, the above two distances are equal.
• That is., P₂P’₂ = FP₂
    ♦ In the fig., both distances are 6.58 units 
3. Consider the point P₃ in the fig.11.15.
This point P₃ satisfies one condition. It can be written in steps:
(i) The distance of P₃ from the magenta line is P₃P’₃
(ii) The distance of P₃ from the point F is FP₃
(iii) The condition is that, the above two distances are equal.
• That is., P₃P’₃ = FP₃
    ♦ In the fig., both distances are 16.19 units 
4. We have seen three points. All of them satisfy the same condition:
    ♦ Distance from the magenta line
    ♦ is equal to
    ♦ Distance from F
5. There are infinite number of points which will satisfy the above condition. All such points will lie on the red curve.
• The red curve is called Parabola.
◼ We can write:
• A parabola is the set of all points that are equidistant from a fixed line and a fixed point.
    ♦ The fixed point must not be on the fixed line.
    ♦ The fixed line, fixed point and all points on the parabola, must lie on the same plane.
        ✰ The fixed line is called the directrix (l) of the parabola.     
        ✰ The fixed point is called the focus (F) of the parabola.
6. We can draw a line in such a way that:
    ♦ The line passes through the focus F
    ♦ The line is perpendicular to the directrix.
• Only one line is possible which satisfies both the conditions. That line is called the axis of the parabola.
• It is shown in fig.11.16 below:

Fig.11.16

7. Consider the point where the axis intersects the parabola. That point is called the vertex (V) of the parabola. It is shown in fig.11.16 above.
8. Consider the point where the axis intersects the directrix. That point is marked as D in the above fig.11.16.
• Let us see the significance of point D. It can be written in four steps:
(i) We know that, all points on the parabola are equidistant from l and F
(ii) V is a point on the parabola. So V is also equidistant from l and F
(iii) That means, V is the midpoint of DF
(iv) In other words:
    ♦ Distance of l  from V
    ♦ is equal to
    ♦ Distance of F from V
9. Let us see how the word parabola is derived:
    ♦ The word ‘para’ means ‘for’
    ♦ The word ‘bola’ means ‘throwing’
• In projectile motion in physics classes, we have seen that, a thrown object follows a parabolic path.

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Standard equations of parabola

• The equation of a parabola will be in the simplest form when two conditions are satisfied:
(i) V is at the origin of the coordinate axes.
(ii) Axis of the parabola coincides with one of the coordinate axes.
• Based on this information, we can think of four cases:
A. V is at the origin, axis coincides with the x-axis and parabola opens to the right.
    ♦ This is shown in fig.11.17(a) below.
        ✰ Axes are shown in cyan color
        ✰ Parabola is shown in red color.
        ✰ Directrix is shown in magenta color. It is marked as l.
        ✰ Axis of the parabola is shown in yellow color.
B. V is at the origin, axis coincides with the x-axis and parabola opens to the left.
    ♦ This is shown in fig.11.17(b) below.
C. V is at the origin, axis coincides with the y-axis and parabola opens upwards.
    ♦ This is shown in fig.11.17(c) below.
D. V is at the origin, axis coincides with the y-axis and parabola opens downwards.
    ♦ This is shown in fig.11.17(d) below.

Fig.11.17

• We will now derive equation in each case.

Case A:
V is at the origin, axis coincides with the x-axis and parabola opens to the right.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.18 below:

Fig.11.18

• Remember the two points:
(i) Axis of the parabola coincides with the x-axis. So F lies on the x-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (a,0)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (-a,0)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘-a’
• P₁ and P lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘y’
• Thus the coordinates of P₁ are: (-a,y)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll}
{}&{PP_1}
&{}={}& {\sqrt{(-a - x)^2~+~(y-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{[-1 × (a + x)]^2~+~(y-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{(a + x)^2}}
&{} \\

\end{array}$

• Next we calculate PF:
$\begin{array}{ll}
{}&{PF}
&{}={}& {\sqrt{(a - x)^2~+~(0-y)^2}}
&{} \\

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

\end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll}
{}&{PP_1}
& {~=~}& {PF}
&{}&{}&{} \\

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

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

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

{\Rightarrow}&{2ax}
& {~=~}& {- 2ax + y^2}
&{}&{}&{} \\

{\Rightarrow}&{y^2}
& {~=~}& {4ax}
&{}&{}&{} \\

\end{array}$

• This is the equation of the parabola for case A.

Case B:
V is at the origin, axis coincides with the x-axis and parabola opens to the left.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.19 below:

Fig.13.19

• Remember the two points:
(i) Axis of the parabola coincides with the x-axis. So F lies on the x-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (-a,0)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (a,0)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘a’
• P₁ and P lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘y’
• Thus the coordinates of P₁ are: (a,y)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll}
{}&{PP_1}
&{}={}& {\sqrt{(a - x)^2~+~(y-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{(a - x)^2}}
&{} \\

\end{array}$

• Next we calculate PF:
$\begin{array}{ll}
{}&{PF}
&{}={}& {\sqrt{(-a - x)^2~+~(0-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{[-1 × (a + x)]^2~+~y^2}}
&{} \\

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

\end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll}
{}&{PP_1}
& {~=~}& {PF}
&{}&{}&{} \\

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

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

{\Rightarrow}&{-2ax}
& {~=~}& {2ax + y^2}
&{}&{}&{} \\

{\Rightarrow}&{y^2}
& {~=~}& {-4ax}
&{}&{}&{} \\

\end{array}$

• This is the equation of the parabola for case B.

Case C:
V is at the origin, axis coincides with the y-axis and parabola opens upwards.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.20 below:

Fig.11.20

• Remember the two points:
(i) Axis of the parabola coincides with the y-axis. So F lies on the y-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (0,a)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (0,-a)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘-a’
• P₁ and P lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘x’
• Thus the coordinates of P₁ are: (x,-a)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll}
{}&{PP_1}
&{}={}& {\sqrt{(x - x)^2~+~(-a-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{[-1 × (a + y)]^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{(a + y)^2}}
&{} \\

\end{array}$

• Next we calculate PF:
$\begin{array}{ll}
{}&{PF}
&{}={}& {\sqrt{(0 - x)^2~+~(a-y)^2}}
&{} \\

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

\end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll}
{}&{PP_1}
& {~=~}& {PF}
&{}&{}&{} \\

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

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

{\Rightarrow}&{2ay}
& {~=~}& {x^2 - 2ay}
&{}&{}&{} \\

{\Rightarrow}&{x^2}
& {~=~}& {4ay}
&{}&{}&{} \\

\end{array}$

• This is the equation of the parabola for case C.

Case D:
V is at the origin, axis coincides with the y-axis and parabola opens downwards.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.21 below:

Fig.11.21

• Remember the two points:
(i) Axis of the parabola coincides with the y-axis. So F lies on the y-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (0,-a)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (0,a)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘a’
• P₁ and P lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘x’
• Thus the coordinates of P₁ are: (x,a)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll}
{}&{PP_1}
&{}={}& {\sqrt{(x - x)^2~+~(a-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{(a - y)^2}}
&{} \\

\end{array}$

• Next we calculate PF:
$\begin{array}{ll}
{}&{PF}
&{}={}& {\sqrt{(0 - x)^2~+~(-a-y)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{x^2~+~[-1 × (a+y)]^2}}
&{} \\

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

\end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll}
{}&{PP_1}
& {~=~}& {PF}
&{}&{}&{} \\

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

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

{\Rightarrow}&{-2ay}
& {~=~}& {x^2 + 2ay}
&{}&{}&{} \\

{\Rightarrow}&{x^2}
& {~=~}& {-4ay}
&{}&{}&{} \\

\end{array}$

• This is the equation of the parabola for case D.

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So we have seen all the four cases. The results follow a pattern. So we can draw a flow chart:

Fig.11.22

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In the next section, we will see latus rectum.

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