Chapter 11.13 - Solved Examples on Hyperbola

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

 

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Link to a few more solved examples is given below:

Exercise 11.4

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In the next section, we will see some miscellaneous examples.

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