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 11.3In the next section, we will see hyperbola.
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