# 12.2 The hyperbola  (Page 4/13)

 Page 6 / 13

Graph the hyperbola given by the equation $\text{\hspace{0.17em}}\frac{{x}^{2}}{144}-\frac{{y}^{2}}{81}=1.\text{\hspace{0.17em}}$ Identify and label the vertices, co-vertices, foci, and asymptotes.

vertices: $\text{\hspace{0.17em}}\left(±12,0\right);\text{\hspace{0.17em}}$ co-vertices: $\text{\hspace{0.17em}}\left(0,±9\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(±15,0\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=±\frac{3}{4}x;$

## Graphing hyperbolas not centered at the origin

Graphing hyperbolas centered at a point $\text{\hspace{0.17em}}\left(h,k\right)$ other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1\text{\hspace{0.17em}}$ for horizontal hyperbolas, and $\text{\hspace{0.17em}}\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1\text{\hspace{0.17em}}$ for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

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

1. Convert the general form to that standard form. 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 center, vertices, co-vertices, foci; and equations for the asymptotes.
1. If the equation is in the form $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ then
• the transverse axis is parallel to the x -axis
• the center is $\text{\hspace{0.17em}}\left(h,k\right)$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h±a,k\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h,k±b\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(h±c,k\right)$
• the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{b}{a}\left(x-h\right)+k$
2. If the equation is in the form $\text{\hspace{0.17em}}\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ then
• the transverse axis is parallel to the y -axis
• the center is $\text{\hspace{0.17em}}\left(h,k\right)$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h,k±a\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h±b,k\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(h,k±c\right)$
• the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{a}{b}\left(x-h\right)+k$
3. Solve for the coordinates of the foci using the equation $\text{\hspace{0.17em}}c=±\sqrt{{a}^{2}+{b}^{2}}.$
4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.

## Graphing a hyperbola centered at ( h , k ) given an equation in general form

Graph the hyperbola    given by the equation $\text{\hspace{0.17em}}9{x}^{2}-4{y}^{2}-36x-40y-388=0.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

$\left(9{x}^{2}-36x\right)-\left(4{y}^{2}+40y\right)=388$

Factor the leading coefficient of each expression.

$9\left({x}^{2}-4x\right)-4\left({y}^{2}+10y\right)=388$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$9\left({x}^{2}-4x+4\right)-4\left({y}^{2}+10y+25\right)=388+36-100$

Rewrite as perfect squares.

$9{\left(x-2\right)}^{2}-4{\left(y+5\right)}^{2}=324$

Divide both sides by the constant term to place the equation in standard form.

$\frac{{\left(x-2\right)}^{2}}{36}-\frac{{\left(y+5\right)}^{2}}{81}=1$

The standard form that applies to the given equation is $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,$ where $\text{\hspace{0.17em}}{a}^{2}=36\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{b}^{2}=81,$ or $\text{\hspace{0.17em}}a=6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b=9.\text{\hspace{0.17em}}$ Thus, the transverse axis is parallel to the x -axis. It follows that:

• the center of the ellipse is $\text{\hspace{0.17em}}\left(h,k\right)=\left(2,-5\right)$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h±a,k\right)=\left(2±6,-5\right),\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(-4,-5\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(8,-5\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h,k±b\right)=\left(2,-5±9\right),\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(2,-14\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,4\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(h±c,k\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c=±\sqrt{{a}^{2}+{b}^{2}}.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,$ we have

$c=±\sqrt{36+81}=±\sqrt{117}=±3\sqrt{13}$

Therefore, the coordinates of the foci are $\text{\hspace{0.17em}}\left(2-3\sqrt{13},-5\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2+3\sqrt{13},-5\right).$

The equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{b}{a}\left(x-h\right)+k=±\frac{3}{2}\left(x-2\right)-5.$

Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in [link] .

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5