In the previous section, we completed a discussion on the standard equations of hyperbola. In this section, we will see latus rectum of hyperbola.
Latus rectum of a hyperbola
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 a hyperbola, it must satisfy three conditions.
(i) It must pass through F1 or F2.
(ii) It must be perpendicular to the transverse axis.
(iii) It’s end points must lie on the hyperbola.
3. Line segments AB and CD in fig.11.52 below satisfy all three conditions. So both are latus rectum of that hyperbola.
 |
| Fig.11.52 |
Length of the latus rectum
•
We have seen the two forms where the equation of hyperbola 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.11.52 above. We want the length AB. It can be calculated in 2 steps:
1. In the fig.11.52 above, let the length AF2 be l.
• Then the coordinates of A will be (c,l)
2. Point A lies on the hyperbola. So we can write:
$\begin{array}{ll}
{}&{\frac{c^2}{a^2}~-~\frac{l^2}{b^2}}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{c^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{a^2~+~b^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{a^2}{a^2}~+~\frac{b^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {1~+~\frac{b^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{b^2}{a^2}}
&{} \\
{\Rightarrow}&{l^2}
& {~=~}& {\frac{b^4}{a^2}}
&{} \\
{\Rightarrow}&{l}
& {~=~}& {\frac{b^2}{a}}
&{} \\
{\Rightarrow}&{2l}
& {~=~}& {\frac{2b^2}{a}}
&{} \\
\end{array}$
• Note:
In the above calculation, first we obtained l. Then we doubled it to obtain the total length AB. This is because, the hyperbola is symmetric about the transverse axis.
◼ So we can write:
Length of the latus rectum of the hyperbola $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$ is $\frac{2 b^2}{a}$
• Let us see Form 2. It is shown in fig.11.53 below:
 |
| Fig.11.53 |
• We want the length AB. It can be calculated in 2 steps:
1. In the fig.11.53 above, let the length AF1 be l.
Then the coordinates of A will be (-l,c)
2. Point A lies on the hyperbola. So we can write:
$\begin{array}{ll}
{}&{\frac{c^2}{a^2}~-~\frac{(-l)^2}{b^2}}
& {~=~}& {1}
&{} \\
{}&{\frac{c^2}{a^2}~-~\frac{l^2}{b^2}}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{c^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {\frac{a^2~+~b^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{\frac{l^2}{b^2}}
& {~=~}& {1~+~\frac{b^2}{a^2}~-~1}
&{} \\
{\Rightarrow}&{l^2}
& {~=~}& {\frac{b^4}{a^2}}
&{} \\
{\Rightarrow}&{l}
& {~=~}& {\frac{b^2}{a}}
&{} \\
{\Rightarrow}&{2l}
& {~=~}& {\frac{2b^2}{a}}
&{} \\
\end{array}$
• Note:
In
the above calculation, first we obtained l. Then we doubled it to
obtain the total length AB. This is because, the hyperbola is symmetric
about the transverse axis.
◼ So we can write:
Length of the latus rectum of the hyperbola $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$ is also $\frac{2 b^2}{a}$
Now we will see a solved example:
Solved example 11.14
For the hyperbolas:
(a) $\frac{x^2}{9}~-~\frac{y^2}{16}~=~1$
(b) y2 - 16x2 = 16
find the following:
(i) coordinates of the foci
(ii) coordinates of the vertices
(iii) the eccentricity
(iv) length of the latus rectum.
Solution:
Part (a):
1. x2 has the +ve term.
• So the equation of the hyperbola is of the form: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
• So we can write:
♦ Transverse axis of this hyperbola lies along the x-axis.
♦ Conjugate axis of this hyperbola lies along the y-axis.
♦ a2 = 9. So a = 3
♦ b2 = 16. So b = 4
2. We have: c2 = a2 + b2.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{c^2}
& {~=~}& {3^2~+~4^2}
&{} \\
{}&{}
& {~=~}& {9~+~16}
&{} \\
{}&{}
& {~=~}& {25}
&{} \\
\end{array}$
• So the value of c is 5
3. The coordinates of the foci are (-c,0) and (c,0)
• So in our present case, the coordinates are: (-5,0) and (5,0)
• 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: (-3,0) and (3,0)
• This is the answer for part (ii).
5. Eccentricity is given by: e = c/a
• So in our present case, e = 5/3
• This is the answer for part (iii).
6. We have: Length of latus rectum = $\frac{2 b^2}{a}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\text{Length}}
& {~=~}& {\frac{2 × 4^2}{3}}
&{} \\
{}&{}
& {~=~}& {\frac{32}{3}~\text{units}}
&{} \\
\end{array}$
• This is the answer for part (iv).
Part (b):
•
The given equation is y2 - 16x2 = 16.
♦ Numerator of the coefficient of x2 must be 1.
♦ Numerator of the coefficient of y2 must be 1.
♦ Right side of the equation must be 1.
•
So we divide the given equation by 16. We get: $\frac{y^2}{16}~-~\frac{x^2}{1}~=~1$
1. y2 has the +ve term.
• So the equation of the hyperbola is of the form: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
• So we can write:
♦ Transverse axis of this hyperbola lies along the y-axis.
♦ Conjugate axis of this hyperbola lies along the x-axis.
♦ a2 = 16. So a = 4
♦ b2 = 1. So b = 1
2. We have: c2 = a2 + b2.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{c^2}
& {~=~}& {4^2~+~^2}
&{} \\
{}&{}
& {~=~}& {16~+~1}
&{} \\
{}&{}
& {~=~}& {17}
&{} \\
\end{array}$
• So the value of c is √17
3. The coordinates of the foci are (0,-c) and (0,c)
• So in our present case, the coordinates are: (0,-√17) and (0,√17)
• 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: (0,-4) and (0,4)
• This is the answer for part (ii).
5. Eccentricity is given by: e = c/a
• So in our present case, e = (√17)/4
• This is the answer for part (iii).
6. We have: Length of latus rectum = $\frac{2 b^2}{a}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\text{Length}}
& {~=~}& {\frac{2 × 1^2}{4}}
&{} \\
{}&{}
& {~=~}& {\frac{1}{2}~\text{units}}
&{} \\
\end{array}$
• This is the answer for part (iv).
In the next section, we will see a few more solved examples.
