Chapter 11.8 - Solved Examples on Ellipse

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

Solved example 11.10
For the ellipse 9x² + 4y² = 36, find the following:
(i) coordinates of the foci
(ii) coordinates of the vertices
(iii) the length of major axis
(iv) the length of minor axis
(v) the eccentricity
(vi) length of the latus rectum.
Solution:
The given equation can be rearranged as follows:

$\begin{array}{ll}
{}&{9x^2 + 4 y^2}
& {~=~}& {36}
&{} \\

{\Rightarrow}&{\frac{9 x^2}{36}~+~\frac{4 y^2}{36}}
& {~=~}& {\frac{36}{36}}
&{} \\

{\Rightarrow}&{\frac{x^2}{36/9}~+~\frac{y^2}{36/4}}
& {~=~}& {\frac{36}{36}}
&{} \\

{\Rightarrow}&{\frac{x^2}{4}~+~\frac{y^2}{9}}
& {~=~}& {1}
&{} \\

\end{array}$

• Now the equation is in the standard form of an ellipse.
1. Comparing the denominators:
    ♦ Denominator of the x² term is 4
    ♦ Denominator of the y² term is 9
• The larger denominator is taken as a²
• So the equation of the ellipse is of the form: $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
• So we can write:
    ♦ Major axis of this ellipse lies along the y-axis.
    ♦ Minor axis of this ellipse lies along the x-axis.
    ♦ a² = 9. So a = 3
    ♦ b² = 4. So b = 2
2. We have: c² = a² - b².
• Substituting the known values, we get:

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

{}&{}
& {~=~}& {9~-~4}
&{} \\

{}&{}
& {~=~}& {5}
&{} \\

\end{array}$

• So the value of c is √5

3. The coordinates of the foci are (0,c) and (0,-c)
• So in our present case, the coordinates are: (0,√5) and (0,-√5)
• This is the answer for part (i).
4. The coordinates of the vertices are (0,a) and (0,-a)
• So in our present case, the coordinates are: (0,3) and (0,-3)
• This is the answer for part (ii).
5. The length of major axis is 2a
• So in our present case, the length is: 2 × 3 = 6 units
• This is the answer for part (iii).
6. The length of minor axis is 2b
• So in our present case, the length is: 2 × 2 = 4 units
• This is the answer for part (iv).
7. Eccentricity is given by: e = c/a
• So in our present case, e = √5/3
• This is the answer for part (v).
8. We have: Length of latus rectum = $\frac{2 b^2}{a}$
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{\text{Length}}
& {~=~}& {\frac{2 × 2^2}{3}}
&{} \\

{}&{}
& {~=~}& {\frac{8}{3}~\text{= 2.7 units}}
&{} \\

\end{array}$

• This is the answer for part (vi).

9. The actual plot is shown below:

Fig.11.41

Solved example 11.11
Find the equation of the ellipse whose vertices are (±13,0) and foci are (±5,0).
Solution:
1. Given that, vertices are (-13,0) and (13,0)
• Vertices are the end points of the major axis. The given vertices lie on the x-axis.
• So we can write:
The major axis of the given ellipse lies along the x-axis.
2. The given vertices are symmetric about the origin O.
• So we can write:
The center of the given ellipse is at O.
3. Based on the above two steps, we can write:
• Equation of the given ellipse is of the form: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
4. The vertices are (-13,0) and (13,0).
• So length of the major axis
= Distance between (-13,0) and (13,0)
= 2a = (13 + 13) = 26 
• Thus we get: a = 13 units.
5. Given that, foci are (-5,0) and (5,0).
• So distance of any one focus from center = c = 5 unit.
6. Now we have 'a' and 'c'. We can calculate 'b'.
• We have: c² = a² - b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{5^2}
& {~=~}& {13^2~-~b^2}
&{} \\

{\Rightarrow}&{25}
& {~=~}& {169~-~b^2}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {169~-~25}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {144}
&{} \\

{\Rightarrow}&{b}
& {~=~}& {12}
&{} \\

\end{array}$

• So the value of b is 12 units.

7. Now we have 'a' and 'b'.
• Based on step (3), we can write:
Equation of the ellipse is: $\frac{x^2}{13^2}~+~\frac{y^2}{12^2}~=~1$

Solved example 11.12
Find the equation of the ellipse whose length of the major axis is 20 and foci are (0,±5)
Solution:
1. Given that, foci are (0,5) and (0,-5)
• Both these points lie on the y-axis.
• So we can write:
Major axis of the ellipse lies along the y-axis.
2. The foci (0,5) and (0,-5) are symmetric about the origin O.
• So we can write:
Center of the given ellipse is at O.
3. Based on the above two steps, we can write:
• Equation of the given ellipse is of the form: $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
4. Given that, foci are (0,5) and (0,-5).
• So distance of any one focus from center = c = 5 unit.
5. Given that, length of the major axis is 20 units.
• So we can write:
Length of the semi major axis = a = 10 units.
6. Now we have 'a' and 'c'. We can calculate 'b'.
• We have: c² = a² - b².
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{5^2}
& {~=~}& {10^2~-~b^2}
&{} \\

{\Rightarrow}&{25}
& {~=~}& {100~-~b^2}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {100~-~25}
&{} \\

{\Rightarrow}&{b^2}
& {~=~}& {75}
&{} \\

{\Rightarrow}&{b}
& {~=~}& {5 \sqrt{3}}
&{} \\

\end{array}$

• So the value of b is $5 \sqrt{3}$ units.

7. Now we have 'a' and 'b'.
• Based on step (3), we can write:
Equation of the ellipse is: $\frac{x^2}{(5 \sqrt{3})^2}~+~\frac{y^2}{10^2}~=~1$
• Which is same as: $\frac{x^2}{75}~+~\frac{y^2}{100}~=~1$

Solved example 11.13
Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4,3) and (-1,4).
Solution:
1. Given that, major axis lies along the x-axis.
• So the equation of the ellipse will be of the form: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
2. Since (4,3) ans (-1,4) lie on the ellipse, we can write:
(i) $\frac{4^2}{a^2}~+~\frac{3^2}{b^2}~=~1$
• Which is same as: $\frac{16}{a^2}~+~\frac{9}{b^2}~=~1$
(ii) $\frac{(-1)^2}{a^2}~+~\frac{4^2}{b^2}~=~1$
• Which is same as: $\frac{1}{a^2}~+~\frac{16}{b^2}~=~1$ 
3. Let $\frac{1}{a^2}~=~P~\text{and}~\frac{1}{b^2}~=~Q$
    ♦ Then 2(i) will become: 16P + 9Q = 1
    ♦ Also, 2(ii) will become: P + 16Q = 1
4. Solving the two equations in step (3), we get:
(i) P = $\frac{1}{a^2}~=~\frac{7}{247}$
• So $a^2 ~=~\frac{247}{7}$
(ii) Q = $\frac{1}{b^2}~=~\frac{15}{247}$
• So $b^2 ~=~\frac{247}{15}$
5. Now we have 'a' and 'b'. So based on step (1), we can write:
Equation of the given ellipse is: $\frac{x^2}{247/7}~+~\frac{y^2}{247/15}~=~1$

---

Link to a few more solved examples is given below:

Exercise 11.3

---

In the next section, we will see hyperbola.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com

Chapter 11.7 - Latus Rectum of Ellipse

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

Latus rectum of a ellipse

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 an ellipse, it must satisfy three conditions.
(i) It must pass through F₁ or F₂.
(ii) It must be perpendicular to the major axis.
(iii) It’s end points must lie on the ellipse.
3. Line segments AB and CD in fig.11.39 below satisfy all three conditions. So both are latus rectum of that ellipse.

Fig.11.39

---

Length of the latus rectum

• We have seen the two forms where the equation of ellipse 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.11.39 above. We want the length AB. It can be calculated in 2 steps:
1. In the fig.11.39 above, let the length AF₂ be l.
• Then the coordinates of A will be (c,l)
2. Point A lies on the ellipse. So we can write:

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

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

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

{\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 ellipse is symmetric about the major axis.

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

---

• Let us see Form 2. It is shown in fig.11.40 below:

Fig.11.40

• We want the length AB. It can be calculated in 2 steps:
1. In the fig.11.40 above, let the length AF₂ be l.
Then the coordinates of A will be (-l,c)
2. Point A lies on the ellipse. So we can write:

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

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

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

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

{\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 ellipse is symmetric about the major axis.

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

---

Now we will see a solved example:

Solved example 11.9
For the ellipse $\frac{x^2}{25}~+~\frac{y^2}{9}~=~1$, find the following:
(i) coordinates of the foci
(ii) coordinates of the vertices
(iii) the length of major axis
(iv) the length of minor axis
(v) the eccentricity
(vi) length of the latus rectum.
Solution:
1. Comparing the denominators:
    ♦ Denominator of the x² term is 25
    ♦ Denominator of the y² term is 9
• The larger denominator is taken as a²
• So the equation of the ellipse is of the form: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
• So we can write:
    ♦ Major axis of this ellipse lies along the x-axis.
    ♦ Minor axis of this ellipse lies along the y-axis.
    ♦ a² = 25. So a = 5
    ♦ b² = 9. So b = 3
2. We have: c² = a² - b².
• Substituting the known values, we get:

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

{}&{}
& {~=~}& {25~-~9}
&{} \\

{}&{}
& {~=~}& {16}
&{} \\

\end{array}$

• So the value of c is 4

3. The coordinates of the foci are (-c,0) and (c,0)
• So in our present case, the coordinates are: (-4,0) and (4,0)
• 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: (-5,0) and (5,0)
• This is the answer for part (ii).
5. The length of major axis is 2a
• So in our present case, the length is: 2 × 5 = 10 units
• This is the answer for part (iii).
6. The length of minor axis is 2b
• So in our present case, the length is: 2 × 3 = 6 units
• This is the answer for part (iv).
7. Eccentricity is given by: e = c/a
• So in our present case, e = 4/5
• This is the answer for part (v).
8. We have: Length of latus rectum = $\frac{2 b^2}{a}$
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{\text{Length}}
& {~=~}& {\frac{2 × 3^2}{5}}
&{} \\

{}&{}
& {~=~}& {\frac{18}{5}~\text{units}}
&{} \\

\end{array}$

• This is the answer for part (vi).

---

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

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com

Chapter 11.6 - Another Simplest Equation of Ellipse

In the previous section, we saw the simplest equation of an ellipse.
    ♦ The major axis was along the x-axis.
    ♦ The minor axis was along the y-axis.
• We will get another simplest equation also when:
    ♦ The major axis lies along the y-axis.
    ♦ The minor axis lies along the x-axis.
• We will see it in this section.

• To write the equation of an ellipse, 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 of the ellipse is at the origin O.
    ♦ The major axis of the ellipse lies along the y-axis.
    ♦ The minor axis of the ellipse lies along the x-axis.
• This is shown in fig.13.36 below:

Fig.11.36

• Based on fig.13.36, we can derive the equation in 6 steps:
1. Let P(x,y) be any point on the ellipse.
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}$

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

5. We know that, B is at a distance of 'a' from O. So the coordinates of B will be (0,-a).
• Now we have three points and their coordinates:
B(0,-a), F₁(0,c), F₂(0,-c)
• Using the distance formula, we can write some distances:

• First we write the distance BF₁:
$\begin{array}{ll}
{}&{BF_1}
& {~=~}& {\sqrt{(0 - 0)^2~+~(-a~-~c)^2}}
&{} \\

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

{}&{}
& {~=~}& {a + c}
&{} \\

\end{array}$ 

• Next we write the distance BF₂:
$\begin{array}{ll}
{}&{BF_2}
& {~=~}& {\sqrt{(0 - 0)^2~+~(-a~-~ -c)^2}}
&{} \\

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

{}&{}
& {~=~}& {a - c}
&{} \\

\end{array}$

• Sum of the above two distances is: (a+c) + (a-c) = 2a

6. Both P and B are points on the same ellipse. So the sum of the distances must be equal.
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{b^2}{a^2} \right)}
& {~=~}& {b^2~~ \color {green} {\text{- - - (V)}}}
&{} \\

{\Rightarrow}&{\frac{x^2}{b^2}~+~\frac{y^2}{a^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, write b² in the place of a² - c².
• Line marked as (V):
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: $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
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~-~ 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}}
& {~=~}& {1-\frac{y^2}{a^2}}
&{} \\

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

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

{}&{}
& {~=~}& {\sqrt{a^2 - y^2 - c^2 + \frac{c^2 y^2}{a^2}~+~y^2 - 2yc + c^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{a^2 + \frac{c^2 y^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 a² - c² 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(1-\frac{y^2}{a^2} \right)~+~(y+c)^2}}
&{} \\

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

{}&{}
& {~=~}& {\sqrt{a^2 - y^2 - c^2 + \frac{c^2 y^2}{a^2}~+~y^2 + 2yc + c^2}}
&{} \\

{}&{}
& {~=~}& {\sqrt{a^2 + \frac{c^2 y^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 a² - c² in the place of b². 

7. So the sum of the distances of P from F₁ and F₂ is:
(Pf₁ + PF₂) = (a - cy/a) + (a + cy/a) = 2a

8. Consider step (5) below fig.11.36 at the beginning of this section. We saw that, sum of the distances of point B from the foci is '2a'.

9. So any point P(x,y) on the ellipse will satisfy the equation $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
• The converse is proved.

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

---

Based on the above equation of the ellipse, we can write two interesting facts:
Fact 1:
This can be written in 5 steps:
1. We have: $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
2. This can be rearranged as: $\frac{x^2}{b^2}~=~1~-~\frac{y^2}{a^2}$
So $\frac{x^2}{b^2}$ will be always less than 1.
That is: $\frac{x^2}{b^2}~\le~1$
⇒ $x^2 ~\le~b^2$
3. Solving the above inequality, we get:
    ♦ x should not be less than -b.
    ♦ x should not be greater than b.
• That is: $-b~\le~x~\le~b$
4. So we can write:
• Consider any point on the ellipse.
    ♦ The x-coordinate of that point will be greater than -b.
    ♦ The x-coordinate of that point will be less than b.
5. So the ellipse will lie between two vertical lines.
    ♦ The left vertical line is x = -b.
    ♦ The right vertical line is x = b.

Fact 2:
This can be written in 5 steps:
1. We have: $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
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 less than 1.
That is: $\frac{y^2}{a^2}~\le~1$
⇒ $y^2 ~\le~a^2$
3. Solving the above inequality, we get:
    ♦ y should not be less than -a.
    ♦ y should not be greater than a.
• That is: $-a~\le~y~\le~a$
4. So we can write:
• Consider any point on the ellipse.
    ♦ The y-coordinate of that point will be greater than -a.
    ♦ The y-coordinate of that point will be less than a.
5. So the ellipse will lie between two horizontal lines.
    ♦ The upper horizontal line is y = a.
    ♦ The lower horizontal line is y = -a.

• The ellipse and the lines in fig.11.37 below, demonstrates the two facts:

Fig.11.37

• Equation of the ellipse in the above fig. is: $\frac{x^2}{3^2}~+~\frac{y^2}{5^2}~=~1$
• We see that:
    ♦ Value of 'a' is 5.
        ✰ The horizontal lines are related to 5.
    ♦ Value of 'b' is 3.
        ✰ The vertical lines are related to 3.

---

• So we have seen the two simplest forms of the ellipse. 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,
    ♦ Major axis lies along the x-axis.
    ♦ Minor axis lies along the y-axis.
2. In the second form,
    ♦ Major axis lies along the y-axis.
    ♦ Minor axis lies along the x-axis.
3. Whatever be the orientation,
    ♦ Major axis is the longer axis.
    ♦ Minor axis is the shorter axis.
4. Whatever be the orientation,
    ♦ Length of major axis is denoted by the letter ‘a’.
    ♦ Length of minor axis is denoted by the letter ‘b
5. 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: $\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1$
• These are known as the standard equations of the ellipse.
B. Comparison of coefficients:
• This can be written in 3 steps:
1. For the first form, $\frac{1}{a^2}$ is the coefficient of x²
2. For the second form, $\frac{1}{a^2}$ is the coefficient of y²
3. We have seen that, ‘a’ is always larger.
◼ So in the given equation,    
• If "denominator of the coefficient" of x² is larger, then the ellipse belongs to the first form.
• If "denominator of the coefficient" of y² is larger, then the ellipse belongs to the second form.
C. Comparison based on symmetry:
• This can be written in 3 steps:
1. Whatever be the orientation, the ellipse will be symmetric about the major axis and minor 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.11.38 below:

Fig.11.38

---

So we have seen the standard equations of the ellipse. In the next section, we will see latus rectum of ellipse.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com

Chapter 11.5 - Simplest Equation of Ellipse

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

• To write the equation of an ellipse, 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 ellipse is at the origin O.
    ♦ The major axis of the ellipse lies along the x-axis.
    ♦ The minor axis of the ellipse lies along the y-axis.
• This is shown in fig.11.34 below:

Fig.11.34

• Based on fig.11.34, we can derive the equation in 6 steps:
1. Let P(x,y) be any point on the ellipse.
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}$

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

5. We know that, B is at a distance of 'a' from O. So the coordinates of B will be (a,0).
• Now we have three points and their coordinates:
B(a,0), F₁(-c,0), F₂(c,0)
• Using the distance formula, we can write some distances:

• First we write the distance BF₁:
$\begin{array}{ll}
{}&{BF_1}
& {~=~}& {\sqrt{(a~-~ -c)^2~+~(0 - 0)^2}}
&{} \\

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

{}&{}
& {~=~}& {a + c}
&{} \\

\end{array}$ 

• Next we write the distance BF₂:
$\begin{array}{ll}
{}&{BF_2}
& {~=~}& {\sqrt{(a~-~ c)^2~+~(0 - 0)^2}}
&{} \\

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

{}&{}
& {~=~}& {a - c}
&{} \\

\end{array}$

• Sum of the above two distances is: (a+c) + (a-c) = 2a

6. Both P and B are points on the same ellipse. So the sum of the distances must be equal.
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}}
& {~=~}& {a~-~\frac{xc}{a}~~ \color {green} {\text{- - - (III)}}}
&{} \\

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

{\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, write b² in the place of a² - c².
• Line marked as (V):
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 9 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}}
& {~=~}& {1-\frac{x^2}{a^2}}
&{} \\

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

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

{}&{}
& {~=~}& {\sqrt{x^2 + 2cx + c^2~+~a^2 - x^2 - c^2 + \frac{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 a² - c² 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(1-\frac{x^2}{a^2} \right)}}
&{} \\

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

{}&{}
& {~=~}& {\sqrt{x^2 - 2cx + c^2~+~a^2 - x^2 - c^2 + \frac{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 a² - c² in the place of b². 

7. So the sum of the distances of P from F₁ and F₂ is:
(PF₁ + PF₂) = (a + xc/a) + (a - xc/a) = 2a.

8. Consider step (5) below fig.11.34 at the beginning of this section. We saw that, sum of the distances of point B from the foci is '2a'.

9. So any point P(x,y) on the ellipse 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 ellipse is at O, major axis lies along the x-axis and minor axis lies along the y-axis, then equation of the ellipse is: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$

---

Based on the above equation of the ellipse, we can write two interesting facts:
Fact 1:
This can be written in 5 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 less than 1.
• That is: $\frac{x^2}{a^2}~\le~1$
⇒ $x^2 ~\le~a^2$
3. Solving the above inequality, we get:
    ♦ x should not be less than -a.
    ♦ x should not be greater than a.
• That is: $-a~\le~x~\le~a$
4. So we can write:
• Consider any point on the ellipse.
    ♦ The x-coordinate of that point will be greater than -a.
    ♦ The x-coordinate of that point will be less than a.
5. So the ellipse will lie between two vertical lines.
    ♦ The left vertical line is x = -a.
    ♦ The right vertical line is x = a.

Fact 2:
This can be written in 5 steps:
1. We have: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
2. This can be rearranged as: $\frac{y^2}{b^2}~=~1~-~\frac{x^2}{a^2}$
• So $\frac{y^2}{b^2}$ will be always less than 1.
• That is: $\frac{y^2}{b^2}~\le~1$
⇒ $y^2 ~\le~b^2$
3. Solving the above inequality, we get:
    ♦ y should not be less than -b.
    ♦ y should not be greater than b.
• That is: $-b~\le~y~\le~a$
4. So we can write:
• Consider any point on the ellipse.
    ♦ The y-coordinate of that point will be greater than -b.
    ♦ The y-coordinate of that point will be less than b.
5. So the ellipse will lie between two horizontal lines.
    ♦ The upper horizontal line is y = b.
    ♦ The lower horizontal line is y = -b.

• The ellipse and the lines in fig.11.35 below, demonstrates the two facts:

Fig.11.35

• Equation of the ellipse in the above fig. is: $\frac{x^2}{5^2}~+~\frac{y^2}{3^2}~=~1$
• We see that:
    ♦ Value of 'a' is 5.
        ✰ The vertical lines are related to 5.
    ♦ Value of 'b' is 3.
        ✰ The horizontal lines are related to 3.

---

• In this section, we saw a simplest equation of an ellipse.
    ♦ The major axis lies along the x-axis.
    ♦ The minor axis lies along the y-axis.
• We will get another simplest equation also when:
    ♦ The major axis lies along the y-axis.
    ♦ The minor axis lies along the x-axis.
• We will see it in the next section.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com

Chapter 11.4 - Ellipse

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

Some basics about ellipse can be written in 6 steps:
1. Consider the five points F₁, F₂, P₁, P₂ and P₃ marked in fig.11.29 below:

Fig.11.29

2. Let us write some distances:
• Distance of P₁:
   ♦ Distance of P₁ from F₁ is 9.8 units.
   ♦ Distance of P₁ from F₂ is 21 units.
         ✰ The sum of the two distances is (9.8+21) = 30.8 units.
• Distance of P₂:
   ♦ Distance of P₂ from F₁ is 24.5 units.
   ♦ Distance of P₂ from F₂ is 6.3 units.
         ✰ The sum of the two distances is (24.5+6.3) = 30.8 units.
• Distance of P₃:
   ♦ Distance of P₃ from F₁ is 11.9 units.
   ♦ Distance of P₃ from F₂ is 18.9 units.
         ✰ The sum of the two distances is (11.9+18.9) = 30.8 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 sum of the two distances will be 30.8 units.
4. There are infinite number of points for which the sum is 30.8 units.      
• All such points will lie in the red curve.
• The red curve is called an ellipse.
5. So we can write the definition:
An ellipse is the set of all points in a plane sum of whose distances from two fixed points in the plane is a constant.
6. Note that, for a particular ellipse, the two points F₁ and F₂ are fixed. If we change one or both of those points, we will get another ellipse.

---

Now we will see some basic features of ellipse. They can be written in 7 steps:
1. We have seen that, F₁ and F₂ are two fixed points.
• They are called the foci of the ellipse.
(‘foci’ is the plural of ‘focus’)
2. Draw a line connecting the two foci. Extend this line in both directions.
• Let it intersect the ellipse at A and B.
• Then the line segment AB is called the major axis of the ellipse.
• This is shown in fig.11.30 below:

Fig.11.30

3. The midpoint of F₁F₂ is called center of the ellipse. It is denoted by the letter O.
4. Draw a line through O and perpendicular to AB.
• Let this line intersect the ellipse at C and D
• Then the line segment CD is called the minor axis of the ellipse.
5. The end points A and B of the major axis are called vertices of the ellipse.
6. The length of the major axis is written as ‘2a’. This is shown in fig.11.31 below:

Fig.13.31

• The length of the minor axis is written as ‘2b’
• The distance between the two foci is written as ‘2c’
7. Based on the above fig.11.31, we can write:
   ♦ Length of semi major axis is ‘a’.   
   ♦ Length of semi minor axis is ‘b’.    
   ♦ Distance from center to any one focus is 'c'.

---

Now we will derive the relation between 'a', 'b' and 'c'. It can be derived in four steps:
1. In fig.12.32 below, B and C are two points on the ellipse.
2. Let us write the distances related to B:
• Distance of B from F₁ can be written as:

$\begin{array}{ll}
{}&{BF_1}
& {~=~}& {OF_1}
&{~+~}&{OF_2}&{~+~}&{B F_2} \\

{}&{}
& {~=~}& {c}
&{~+~}&{c}&{~+~}&{OB - OF_2} \\

{}&{}
& {~=~}& {c}
&{~+~}&{c}&{~+~}&{a - c} \\

{}&{}
& {~=~}& {c}
&{}&{}&{~+~}&{a} \\

\end{array}$

• Distance of B from F₂ can be written as:

$\begin{array}{ll}
{}&{BF_2}
& {~=~}& {OB}
&{~-~}&{OF_2}&{}&{} \\

{}&{}
& {~=~}& {a}
&{~-~}&{c}&{}&{} \\

\end{array}$

• So sum of the two distances = (c+a) + (a-c) = 2a

3. Let us write the distances related to C.
• Distance of C from F₁ can be written as:

$\begin{array}{ll}
{}&{CF_1}
& {~=~}& {\sqrt{(O F_1)^2 ~+~(OC)^2}}
&{}&{}&{}&{} \\

{}&{}
& {~=~}& {\sqrt{c^2 ~+~b^2}}
&{}&{}&{}&{} \\

\end{array}$

• Distance of C from F₂ can be written as:

$\begin{array}{ll}
{}&{CF_2}
& {~=~}& {\sqrt{(O F_2)^2 ~+~(OC)^2}}
&{}&{}&{}&{} \\

{}&{}
& {~=~}& {\sqrt{c^2 ~+~b^2}}
&{}&{}&{}&{} \\

\end{array}$

• So sum of the two distances
= $\sqrt{c^2 ~+~b^2} + \sqrt{c^2 ~+~b^2}~=~2 \sqrt{c^2 ~+~b^2}$

4. Both B and C are points on the same ellipse. So we can equate the sum of the distances. We get:

$\begin{array}{ll}
{}&{2a}
& {~=~}& {2\sqrt{c^2 ~+~b^2}}
&{}&{}&{}&{} \\

{\Rightarrow}&{a}
& {~=~}& {\sqrt{c^2 ~+~b^2}}
&{}&{}&{}&{} \\

{\Rightarrow}&{a^2}
& {~=~}& {b^2 ~+~c^2}
&{}&{}&{}&{} \\

\end{array}$

• This is a simple relation between 'a', 'b' and 'c'.

---

The relation a² = b² + c² is applicable to any ellipse. Let us see two cases where this relation gives interesting results.

Case 1: Special case of an ellipse where it becomes a circle.
This can be written in five steps:
1. Suppose that, the vertices A and B are fixed in position.
    ♦ When A and B are fixed, the length 2a becomes a constant.
    ♦ When 2a is a constant, ‘a’ is a constant.
2. Keeping ‘a’ constant, we decrease ‘c’.
• We have the relation: a² = b² + c².
• So, keeping ‘a’ constant, if we decrease ‘c’, The length ‘b’ will automatically increase.
3. An increase in ‘b’ is an increase in the length of minor axis.
• As the length of the minor axis increase, the shape of the ellipse will become more and more circular.
• This is shown in the animation below:

Fig.13.32

4. We are decreasing ‘c’.
• That means, the foci get closer and closer to O.
• Finally, when ‘c’ becomes zero, the foci will merge at the center O.
• In such a situation, ‘a’ and ‘b’ are equal. The ellipse has become a circle.
5. So we can write:
A circle is an ellipse with a = b and c = 0.

Case 2: Special case of an ellipse where it becomes a line.
This can be written in five steps:
1. Suppose that, the vertices A and B are fixed in position.
    ♦ When A and B are fixed, the length 2a becomes a constant.
    ♦ When 2a is a constant, ‘a’ is a constant.
2. Keeping ‘a’ constant, we increase ‘c’.
• We have the relation: a² = b² + c².
• So, keeping ‘a’ constant, if we increase ‘c’, The length ‘b’ will automatically decrease.
3. A decrease in ‘b’ is a decrease in the length of minor axis.
• As the length of the minor axis decrease, the shape of the ellipse will become more and more linear.
• This is shown in the animation below:

Fig.11.33

4. We are increasing ‘c’.
• That means, the foci get closer and closer to vertices A and B.
• Finally, when ‘c’ becomes 'a', the foci will merge with A and B.
• In such a situation, ‘b’ is zero. The ellipse has become a line.
5. So we can write:
A line is an ellipse with a = c and b = 0.

---

Eccentricity of an Ellipse

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 an ellipse 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 ellipse, the focus is at a distance of ae from the center.
5. We have seen that, for a circle, a = c
• So for a circle, c = ce, which gives e = 1
• We can write:
For a circle, eccentricity is 1

---

In the next section, we will see the standard equations of an ellipse.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com