# 12.2 The hyperbola  (Page 3/13)

 Page 5 / 13

## Finding the equation of a hyperbola centered at ( h , k ) given its foci and vertices

What is the standard form equation of the hyperbola    that has vertices at $\left(0,-2\right)$ and $\left(6,-2\right)$ and foci at $\left(-2,-2\right)$ and $\left(8,-2\right)?$

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

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

First, we identify the center, $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ The center is halfway between the vertices $\text{\hspace{0.17em}}\left(0,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(6,-2\right).\text{\hspace{0.17em}}$ Applying the midpoint formula, we have

$\left(h,k\right)=\left(\frac{0+6}{2},\frac{-2+\left(-2\right)}{2}\right)=\left(3,-2\right)$

Next, we find $\text{\hspace{0.17em}}{a}^{2}.\text{\hspace{0.17em}}$ The length of the transverse axis, $\text{\hspace{0.17em}}2a,$ is bounded by the vertices. So, we can find $\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ by finding the distance between the x -coordinates of the vertices.

Now we need to find $\text{\hspace{0.17em}}{c}^{2}.\text{\hspace{0.17em}}$ The coordinates of the foci are $\text{\hspace{0.17em}}\left(h±c,k\right).\text{\hspace{0.17em}}$ So $\text{\hspace{0.17em}}\left(h-c,k\right)=\left(-2,-2\right)\text{\hspace{0.17em}}$ and $\left(h+c,k\right)=\left(8,-2\right).\text{\hspace{0.17em}}$ We can use the x -coordinate from either of these points to solve for $\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$ Using the point $\left(8,-2\right),\text{\hspace{0.17em}}$ and substituting $\text{\hspace{0.17em}}h=3,$

Next, solve for $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}:$

Finally, substitute the values found for $\text{\hspace{0.17em}}h,k,{a}^{2},$ and $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ into the standard form of the equation.

$\text{\hspace{0.17em}}\frac{{\left(x-3\right)}^{2}}{9}-\frac{{\left(y+2\right)}^{2}}{16}=1$

What is the standard form equation of the hyperbola that has vertices $\text{\hspace{0.17em}}\left(1,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,\text{8}\right)\text{\hspace{0.17em}}$ and foci $\text{\hspace{0.17em}}\left(1,-10\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,16\right)?$

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

## Graphing hyperbolas centered at the origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\text{\hspace{0.17em}}$ for horizontal hyperbolas and the standard form $\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1\text{\hspace{0.17em}}$ for vertical hyperbolas.

Given a standard form equation for a hyperbola centered at $\text{\hspace{0.17em}}\left(0,0\right),\text{\hspace{0.17em}}$ sketch the graph.

1. Determine which of the standard forms applies to the given equation.
2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
1. If the equation is in the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ then
• the transverse axis is on the x -axis
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(±a,0\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(0,±b\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(±c,0\right)$
• the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{b}{a}x$
2. If the equation is in the form $\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1,$ then
• the transverse axis is on the y -axis
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(0,±a\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(±b,0\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±c\right)$
• the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{a}{b}x$
3. Solve for the coordinates of the foci using the equation $\text{\hspace{0.17em}}c=±\sqrt{{a}^{2}+{b}^{2}}.$
4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.

## Graphing a hyperbola centered at (0, 0) given an equation in standard form

Graph the hyperbola    given by the equation $\text{\hspace{0.17em}}\frac{{y}^{2}}{64}-\frac{{x}^{2}}{36}=1.\text{\hspace{0.17em}}$ Identify and label the vertices, co-vertices, foci, and asymptotes.

The standard form that applies to the given equation is $\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1.\text{\hspace{0.17em}}$ Thus, the transverse axis is on the y -axis

The coordinates of the vertices are $\text{\hspace{0.17em}}\left(0,±a\right)=\left(0,±\sqrt{64}\right)=\left(0,±8\right)$

The coordinates of the co-vertices are

The coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±c\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c=±\sqrt{{a}^{2}+{b}^{2}}.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,\text{\hspace{0.17em}}$ we have

$c=±\sqrt{{a}^{2}+{b}^{2}}=±\sqrt{64+36}=±\sqrt{100}=±10$

Therefore, the coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±10\right)$

The equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{a}{b}x=±\frac{8}{6}x=±\frac{4}{3}x$

Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in [link] .

find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
Give me the reciprocal of even number
Aliyu
The reciprocal of an even number is a proper fraction
Jamilu
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve