Hyperbolas

Determine the equation of the hyperbola represented by the graph below.

Solution

The hyperbola below has foci at (0 , –5) and (0, 5) while the vertices are located at (0, –4) and (0, 4). The distance between these two coordinates is 8 units.

Thus, the difference between the distance from any point (x, y) on the hyperbola to the foci is 8 or –8 units, depending on the order in which you subtract.

Using the Distance Formula, we obtain the equation of the hyperbola as follows.

Let,

d₁ = distance from (0, 5) to (x, y)

d₂ = distance from (0, –5) to (x, y)

$$\begin{gathered} d_2-d_1=\pm 8 \\ \Rightarrow \sqrt{{(y-5)}^2+x^2}-\sqrt{{(y+5)}^2+x^2}=\pm 8 \\ \Rightarrow \sqrt{{(y-5)}^2+x^2}=\pm 8+\sqrt{{(y+5)}^2+x^2} \\ \Rightarrow {(y-5)}^2+x^2={\left(\pm 8+\sqrt{{(y+5)}^2+x^2}\right)}^2 \\ \Rightarrow y^2-10y+25+x^2=64+16\sqrt{{(y+5)}^2+x^2}+{(y+5)}^2+x^2 \\ \Rightarrow \cancel{y^2}-10y+\cancel{25}+\cancel{x^2}=64\pm 16\sqrt{{(y+5)}^2+x^2}+\cancel{y^2}+10y+\cancel{25}+\cancel{x^2} \\ \Rightarrow -20y-64=\pm 16\sqrt{{(y+5)}^2+x^2} \\ \Rightarrow 5y+16=\pm 4\sqrt{{(y+5)}^2+x^2} \\ \Rightarrow {\left(5y+16\right)}^2=16{\left(\sqrt{{(y+5)}^2+x^2}\right)}^2 \\ \Rightarrow 25y^2+160y+256=16(y^2+10y+25+x^2) \\ \Rightarrow 25y^2+\cancel{160y}+256=16y^2+\cancel{160y}+400+16x^2 \\ \Rightarrow 9y^2-16x^2=144 \end{gathered}$$

By dividing the expression by $9\times 16,$ our final yield becomes

$\Rightarrow \frac{y^2}{16}+\frac{x^2}{9}=1$

Identify the foci and vertices for the hyperbola $\frac{y^2}{49}-\frac{x^2}{32}=1$.

Solution

The equation is of the form

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Thus, the transverse axis lies on the y-axis. The centre is at the origin, so the y-intercepts are the vertices of the graph. We can thus find the vertices by setting x = 0 and solve for y as below.

$$\begin{gathered} \frac{y^2}{49}-\frac{{\left(0\right)}^2}{32}=1 \\ \Rightarrow \frac{y^2}{49}=1 \\ \Rightarrow y^2=49 \\ \Rightarrow y=\pm \sqrt{49}=\pm 7 \end{gathered}$$

The foci are located at $(0,\pm c)$ and by the relationship between a, b and c established before, we obtain

$c^2=a^2+b^2\Rightarrow c=\pm \sqrt{a^2+b^2}=\pm \sqrt{49+32}=\pm \sqrt{81}=\pm 9$

Thus, the vertices are (0, –7) and (0, 7) while the foci are (0, –9) and (0, 9). The graph is shown below.

Identify the foci and vertices for the hyperbola $\frac{{(y-2)}^2}{9}-\frac{{(x-3)}^2}{25}=1$.

Solution

The equation is of the form

$\frac{{(y-k)}^2}{a^2}-\frac{{(x-h)}^2}{b^2}=1$

Thus, the transverse axis lies on the y-axis. Here, h = 3 and k = 2, so the centre is at (3, 2). To find the vertices, we shall make use of the formula below.

$(h,k\pm a)\Rightarrow (3,2\pm \sqrt{9})=(3,2\pm 3)$

The foci are located at $(0,\pm c)$ and using $c^2=a^2+b^2$ as before, we obtain

$c=\pm \sqrt{9+25}=\pm \sqrt{34}\approx \pm 5.83(correcttotwodecimalplaces)$

Thus, the vertices are (3, –1) and (3, 5) while the foci are (0, –5.83) and (0, 5.83). The graph is shown below.

Express the following hyperbola in standard form given the following foci and vertices.

$$\begin{gathered} Vertices:(\pm 6,0) \\ Foci:(\pm 2\sqrt{10},0) \end{gathered}$$

Solution

Notice that the vertices and foci are on the x-axis. Hence, the equation of the hyperbola will take the form

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Since the vertices are $(\pm 6,0),$then

$a=6\Rightarrow a^2=36$

Given the foci are $(\pm 2\sqrt{10},0),$then

Solving for b² we obtain

$$\begin{gathered} b^2=c^2-a^2=40-36 \\ \Rightarrow b^2=4 \end{gathered}$$

Now that we have found a² and b², we can substitute this into the standard form as

$\frac{x^2}{36}-\frac{y^2}{4}=1$

The graph is shown below.

Express the following hyperbola in standard form given the following foci and vertices.

$$\begin{gathered} Vertices:(1,-2)and(1,8) \\ Foci:(1,-10)and(1,16) \end{gathered}$$

Solution

The vertices and foci have the same x-coordinates, so the transverse axis is parallel to the y-axis. The equation of the hyperbola will thus take the form

$\frac{{(y-k)}^2}{a^2}-\frac{{(x-h)}^2}{b^2}=1$

We must first identify the centre using the midpoint formula. The centre lies between the vertices (1, –2) and (1, 8), so

$(h,k)=\left(\frac{1+1}{2},\frac{-2+8}{2}\right)=(1,3)$

The length of the transverse axis, 2a, is bounded by the vertices. To find a² we must evaluate the distance between the y-coordinates of the vertices.

$$\begin{gathered} 2a=\left|-2-8\right|=\left|-10\right|=10 \\ \Rightarrow a=5 \\ \Rightarrow a^2=25 \end{gathered}$$

The coordinates of the foci are

$(h,k\pm c)\Rightarrow (h,k-c)=(1,-10)and(h,k+c)=(1,16)$

Using k + 3 = 16 and substituting k = 3, we obtain

$c=13\Rightarrow c^2=169$

Thus, we can solve for b² by

$b^2=c^2-a^2\Rightarrow b^2=169-25=144$

Finally, substituting these values into the standard form, we obtain

$\frac{{(y-3)}^2}{25}-\frac{{(x-1)}^2}{144}=1$

The graph is shown below.

Graph the hyperbola $\frac{x^2}{36}-\frac{y^2}{4}=1$

Solution

As you can see, the hyperbola is already in the standard form.

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

This means that we have a pair of curves opening from the left and right. The vertices are $(\pm 6,0)$ and the foci are $(\pm 2\sqrt{10},0)$. The centre of the hyperbola is the origin, (0, 0). Here,

$$\begin{gathered} a^2=36\Rightarrow a=6 \\ and{b}^2=4\Rightarrow b=2 \end{gathered}$$

The equation of the asymptotes are

$y=\pm \frac{b}{a}x=\pm \frac{2}{6}x=\pm \frac{1}{3}x$

Always sketch the asymptotes when graphing hyperbolas. That way, we can accurately draw the curves associated with the equation.

The graph of $\frac{x^2}{36}-\frac{y^2}{4}=1$ is shown below.

Graph the hyperbola below.

Solution

To solve this expression, we must attempt to rearrange this in the standard form of a hyperbola. We do this by completing the square as follows.

$$\begin{gathered} 9x^2-36x-25y^2-50y=214 \\ 9(x^2-4x+\mathit{A})-25(y^2+2y+\mathit{B})=214+9\left(\mathit{A}\right)-25\left(\mathit{B}\right) \end{gathered}$$

We need to identify A and B. In doing so, we obtain

$$\begin{gathered} 9(x^2-4x+\mathbf{4})-25(y^2+2y+\mathbf{1})=214+9\left(\mathbf{4}\right)-25\left(\mathbf{1}\right) \\ \Rightarrow 9{(x-2)}^2-25{(y+1)}^2=225 \end{gathered}$$

Dividing both sides by 225, we obtain the equation

$\Rightarrow \frac{{(x-2)}^2}{25}-\frac{{(y+1)}^2}{9}=1$

The centre is (2, –1). We also have the values $h=2,k=-1,a=5,b=3andc=\sqrt{34}$. Thus we obtain the following values for the vertices, foci and asymptotes.

The vertices

$$\begin{gathered} (h\pm a,k)=(2\pm 5,-1) \\ \Rightarrow (-3,-1)and(7,-1) \end{gathered}$$

The foci

$$\begin{gathered} (h\pm c,k)=(2\pm \sqrt{34},-1) \\ \Rightarrow (2-\sqrt{34},-1)and(2-\sqrt{34},-1) \end{gathered}$$

The asymptotes

$$\begin{gathered} y=k\pm \frac{b}{a}(x-h)=-1\pm \frac{3}{5}(x-2) \\ \Rightarrow y+1=-\frac{3}{5}(x-2)andy+1=\frac{3}{5}(x-2) \end{gathered}$$

The graph of $\frac{{(x-2)}^2}{25}-\frac{{(y+1)}^2}{9}=1$ is shown below.

Find the eccentricity of the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$

Solution

Here, a² = 25 and b² = 9. Thus, the eccentricity of this hyperbola is given by

$$\begin{gathered} a=\sqrt{25}=5 \\ \Rightarrow e=\frac{\sqrt{25+9}}{5} \\ \Rightarrow e=\frac{\sqrt{34}}{5}\approx 1.17(correcttotwodecimalplaces) \end{gathered}$$