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.
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