1.5 Conic sections  (Page 6/23)

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$\begin{array}{ccc}\hfill -2cx& =\hfill & 4{a}^{2}+4a\sqrt{{\left(x+c\right)}^{2}+{y}^{2}}+2cx\hfill \\ \hfill 4a\sqrt{{\left(x+c\right)}^{2}+{y}^{2}}& =\hfill & -4{a}^{2}-4cx\hfill \\ \hfill \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}& =\hfill & \text{−}a-\frac{cx}{a}\hfill \\ \hfill {\left(x+c\right)}^{2}+{y}^{2}& =\hfill & {a}^{2}+2cx+\frac{{c}^{2}{x}^{2}}{{a}^{2}}\hfill \\ \hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =\hfill & {a}^{2}+2cx+\frac{{c}^{2}{x}^{2}}{{a}^{2}}\hfill \\ \hfill {x}^{2}+{c}^{2}+{y}^{2}& =\hfill & {a}^{2}+\frac{{c}^{2}{x}^{2}}{{a}^{2}}.\hfill \end{array}$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$\begin{array}{}\\ \hfill {x}^{2}-\frac{{c}^{2}{x}^{2}}{{a}^{2}}+{y}^{2}& =\hfill & {a}^{2}-{c}^{2}\hfill \\ \hfill \frac{\left({a}^{2}-{c}^{2}\right){x}^{2}}{{a}^{2}}+{y}^{2}& =\hfill & {a}^{2}-{c}^{2}.\hfill \end{array}$

Finally, divide both sides by ${a}^{2}-{c}^{2}.$ This gives the equation

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}-{c}^{2}}=1.$

We now define b so that ${b}^{2}={c}^{2}-{a}^{2}.$ This is possible because $c>a.$ Therefore the equation of the ellipse becomes

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

Finally, if the center of the hyperbola is moved from the origin to the point $\left(h,k\right),$ we have the following standard form of a hyperbola.

Equation of a hyperbola in standard form

Consider the hyperbola with center $\left(h,k\right),$ a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is

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

and the foci are located at $\left(h±c,k\right),$ where ${c}^{2}={a}^{2}+{b}^{2}.$ The equations of the asymptotes are given by $y=k±\frac{b}{a}\left(x-h\right).$ The equations of the directrices are

$x=k±\frac{{a}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}=h±\frac{{a}^{2}}{c}.$

If the major axis is vertical, then the equation of the hyperbola becomes

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

and the foci are located at $\left(h,k±c\right),$ where ${c}^{2}={a}^{2}+{b}^{2}.$ The equations of the asymptotes are given by $y=k±\frac{a}{b}\left(x-h\right).$ The equations of the directrices are

$y=k±\frac{{a}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}=k±\frac{{a}^{2}}{c}.$

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form $A{x}^{2}+B{y}^{2}+Cx+Dy+E=0,$ where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Finding the standard form of a hyperbola

Put the equation $9{x}^{2}-16{y}^{2}+36x+32y-124=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

First add 124 to both sides of the equation:

$9{x}^{2}-16{y}^{2}+36x+32y=124.$

Next group the x terms together and the y terms together, then factor out the common factors:

$\begin{array}{ccc}\hfill \left(9{x}^{2}+36x\right)-\left(16{y}^{2}-32y\right)& =\hfill & 124\hfill \\ \hfill 9\left({x}^{2}+4x\right)-16\left({y}^{2}-2y\right)& =\hfill & 124.\hfill \end{array}$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ${\left(\frac{4}{2}\right)}^{2}=4.$ In the second set of parentheses, take half the coefficient of y and square it. This gives ${\left(\frac{-2}{2}\right)}^{2}=1.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

$\begin{array}{}\\ 9\left({x}^{2}+4x+4\right)-16\left({y}^{2}-2y+1\right)=124+36-16\hfill \\ 9\left({x}^{2}+4x+4\right)-16\left({y}^{2}-2y+1\right)=144.\hfill \end{array}$

Next factor both sets of parentheses and divide by 144:

$\begin{array}{}\\ \hfill 9{\left(x+2\right)}^{2}-16{\left(y-1\right)}^{2}& =\hfill & 144\hfill \\ \hfill \frac{9{\left(x+2\right)}^{2}}{144}-\frac{16{\left(y-1\right)}^{2}}{144}& =\hfill & 1\hfill \\ \hfill \frac{{\left(x+2\right)}^{2}}{16}-\frac{{\left(y-1\right)}^{2}}{9}& =\hfill & 1.\hfill \end{array}$

The equation is now in standard form. Comparing this to [link] gives $h=-2,$ $k=1,$ $a=4,$ and $b=3.$ This is a horizontal hyperbola with center at $\left(-2,1\right)$ and asymptotes given by the equations $y=1±\frac{3}{4}\left(x+2\right).$ The graph of this hyperbola appears in the following figure.

Put the equation $4{y}^{2}-9{x}^{2}+16y+18x-29=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

$\frac{{\left(y+2\right)}^{2}}{9}-\frac{{\left(x-1\right)}^{2}}{4}=1.$ This is a vertical hyperbola. Asymptotes $y=-2±\frac{3}{2}\left(x-1\right).$ Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
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