In the previous section, we saw latus rectum of hyperbola. We also saw a solved example. In this section, we will see a few more solved examples.
Solved example 11.15
Find the equation of the hyperbola with foci (0,±3) and vertices $\left(0,\pm \frac{\sqrt{11}}{2} \right)$
Solution:
1. The given foci are: (0,3) and (0,-3)
♦ These foci lie on the y-axis.
•
So the equation of the hyperbola is of the form: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
2. Since the foci are (0,3) and (0,-3), we get: c = 3
3. The given vertices are: $\left(0, \frac{\sqrt{11}}{2} \right),~\left(0, -\frac{\sqrt{11}}{2} \right)$
•
So we get: a = $\frac{\sqrt{11}}{2}$
4. Now we have 'a' and 'c'. We can calculate 'b'.
• We have: c2 = a2 + b2.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{3^2}
& {~=~}& {\left(\frac{\sqrt{11}}{2} \right)^2~+~b^2}
&{} \\
{\Rightarrow}&{9}
& {~=~}& {\frac{11}{4}~+~b^2}
&{} \\
{\Rightarrow}&{b^2}
& {~=~}& {9~-~\frac{11}{4}}
&{} \\
{\Rightarrow}&{b^2}
& {~=~}& {\frac{25}{4}}
&{} \\
{\Rightarrow}&{b}
& {~=~}& {\frac{5}{2}}
&{} \\
\end{array}$
• So the value of b is $\frac{5}{2}$ units.
5. Now we have 'a' and 'b'.
• Based on step (1), we can write:
Equation of the hyperbola is: $\frac{y^2}{\left(\frac{\sqrt{11}}{2} \right)^2}~-~\frac{x^2}{\left(\frac{5}{2} \right)^2}~=~1$
•
This can be simplified as follows:
$\begin{array}{ll}
{}&{\frac{y^2}{\left(\frac{\sqrt{11}}{2} \right)^2}~-~\frac{x^2}{\left(\frac{5}{2} \right)^2}}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{\frac{y^2}{\frac{11}{4}}~-~\frac{x^2}{\frac{25}{4}}}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{\frac{4 y^2}{11}~-~\frac{4 x^2}{25}}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{100 y^2~-~44 x^2}
& {~=~}& {275}
&{} \\
\end{array}$
Solved example 11.16
Find the equation of the hyperbola with foci (0,±12) and length of latus rectum 36.
Solution:
1. The given foci are: (0,12) and (0,-12)
♦ These foci lie on the y-axis.
•
So the equation of the hyperbola is of the form: $\frac{y^2}{a^2}~-~\frac{x^2}{b^2}~=~1$
2. Since the foci are (0,12) and (0,-12), we get: c = 12
3. We have: Length of latus rectum = $\frac{2 b^2}{a}$ = 36
• From this, we get:
$\begin{array}{ll}
{}&{36}
& {~=~}& {\frac{2 × b^2}{a}}
&{} \\
{\Rightarrow}&{18}
& {~=~}& {\frac{b^2}{a}}
&{} \\
{\Rightarrow}&{b^2}
& {~=~}& {18a}
&{} \\
\end{array}$
4. Now we have 'b2' in terms of 'a' . We can calculate 'a'.
• We have: c2 = a2 + b2.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{12^2}
& {~=~}& {a^2~+~18 a}
&{} \\
{\Rightarrow}&{144}
& {~=~}& {a^2~+~18a}
&{} \\
{\Rightarrow}&{a^2~+~18a~-~144}
& {~=~}& {0}
&{} \\
\end{array}$
• Solving this quadratic equation, we get: a = 6 or a = -24.
'a' is a length. It cannot be -ve. So we can write: a = 6
5. From the result in (3), we get: b2 = 18a = 18 × 6 = 108
6. Now we have 'a' and 'b2'.
• Based on step (1), we can write:
Equation of the hyperbola is: $\frac{y^2}{36}~-~\frac{x^2}{108}~=~1$
Solved example 11.17
If the foci of the ellipse $\frac{x^2}{25}~+~\frac{y^2}{k^2}~=~1$ and the hyperbola $\frac{x^2}{144}~-~\frac{y^2}{81}~=~\frac{1}{25}$ coincide, find the value of k.
Solution:
1. We are given the complete equation of the hyperbola. So we will analyze it first.
•
Given equation of the hyperbola is: $\frac{x^2}{144}~-~\frac{y^2}{81}~=~\frac{1}{25}$.
•
It can be rearranged as follows:
$\begin{array}{ll}
{}&{\frac{x^2}{144}~-~\frac{y^2}{81}}
& {~=~}& {\frac{1}{25}}
&{} \\
{\Rightarrow}&{\frac{25 x^2}{144}~-~\frac{25 y^2}{81}}
& {~=~}& {\frac{25}{25}}
&{} \\
{\Rightarrow}&{\frac{x^2}{144/25}~-~\frac{y^2}{81/25}}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{\frac{x^2}{(12/5)^2}~-~\frac{y^2}{(9/5)^2}}
& {~=~}& {1}
&{} \\
\end{array}$
2. In the above result, x2 is the +ve term. So we can write:
The given hyperbola is of the form: $\frac{x^2}{a^2}~-~\frac{y^2}{b^2}~=~1$
•
Thus we get: a = 12/5 and b = 9/5
3. For any hyperbola, we have: c2 = a2 + b2.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{c^2}
& {~=~}& {\left(\frac{12}{5} \right)^2~+~\left(\frac{9}{5} \right)^2}
&{} \\
{}&{}
& {~=~}& {\frac{144}{25}~+~\frac{81}{25}}
&{} \\
{}&{}
& {~=~}& {\frac{225}{25}}
&{} \\
\end{array}$
• So the value of c is 15/5 = 3
4. The foci are (-c,0) and (c,0)
•
So we get: (-3,0) and (3,0)
5. Given that, foci of the ellipse and the hyperbola are the same. So we can write:
•
Foci of the ellipse $\frac{x^2}{25}~+~\frac{y^2}{k^2}~=~1$ are: (-3,0) and (3,0)
•
Thus we can write: Value of 'c' for the ellipse is 3.
6. Given equation of the ellipse is: $\frac{x^2}{25}~+~\frac{y^2}{k^2}~=~1$
•
This is of the form: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
•
So we can write: a = 5 and b = k
7. For any ellipse, we have: c2 = a2 - b2.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{3^2}
& {~=~}& {5^2~-~k^2}
&{} \\
{\Rightarrow}&{k^2}
& {~=~}& {5^2~-~3^2}
&{} \\
{\Rightarrow}&{k^2}
& {~=~}& {25~-~9}
&{} \\
{\Rightarrow}&{k^2}
& {~=~}& {16}
&{} \\
{\Rightarrow}&{k}
& {~=~}& {4}
&{} \\
\end{array}$
8. So we can write:
♦ The ellipse $\frac{x^2}{25}~+~\frac{y^2}{4^2}~=~1$
♦ And the hyperbola $\frac{x^2}{144}~-~\frac{y^2}{81}~=~\frac{1}{25}$
♦ Have the same foci (-3,0) and (3,0).
9. The actual plot is shown in fig.11.54 below:
 |
| Fig.11.54 |
Link to a few more solved examples is given below:
Exercise 11.4In the next section, we will see some miscellaneous examples.
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