12.2 The hyperbola  (Page 2/13)

 Page 2 / 13

Deriving the equation of an ellipse centered at the origin

Let $\text{\hspace{0.17em}}\left(-c,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(c,0\right)\text{\hspace{0.17em}}$ be the foci    of a hyperbola centered at the origin. The hyperbola is the set of all points $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ such that the difference of the distances from $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ to the foci is constant. See [link] .

If $\text{\hspace{0.17em}}\left(a,0\right)\text{\hspace{0.17em}}$ is a vertex of the hyperbola, the distance from $\text{\hspace{0.17em}}\left(-c,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(a,0\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}a-\left(-c\right)=a+c.\text{\hspace{0.17em}}$ The distance from $\text{\hspace{0.17em}}\left(c,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(a,0\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}c-a.\text{\hspace{0.17em}}$ The sum of the distances from the foci to the vertex is

$\left(a+c\right)-\left(c-a\right)=2a$

If $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ is a point on the hyperbola, we can define the following variables:

By definition of a hyperbola, $\text{\hspace{0.17em}}{d}_{2}-{d}_{1}\text{\hspace{0.17em}}$ is constant for any point $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the hyperbola. We know that the difference of these distances is $\text{\hspace{0.17em}}2a\text{\hspace{0.17em}}$ for the vertex $\text{\hspace{0.17em}}\left(a,0\right).\text{\hspace{0.17em}}$ It follows that $\text{\hspace{0.17em}}{d}_{2}-{d}_{1}=2a\text{\hspace{0.17em}}$ for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula    . The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

This equation defines a hyperbola centered at the origin with vertices $\text{\hspace{0.17em}}\left(±a,0\right)\text{\hspace{0.17em}}$ and co-vertices $\text{\hspace{0.17em}}\left(0±b\right).$

Standard forms of the equation of a hyperbola with center (0,0)

The standard form of the equation of a hyperbola with center $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and transverse axis on the x -axis is

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

where

• the length of the transverse axis is $\text{\hspace{0.17em}}2a$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(±a,0\right)$
• the length of the conjugate axis is $\text{\hspace{0.17em}}2b$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(0,±b\right)$
• the distance between the foci is $\text{\hspace{0.17em}}2c,$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(±c,0\right)$
• the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{b}{a}x$

The standard form of the equation of a hyperbola with center $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and transverse axis on the y -axis is

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

where

• the length of the transverse axis is $\text{\hspace{0.17em}}2a$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(0,±a\right)$
• the length of the conjugate axis is $\text{\hspace{0.17em}}2b$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(±b,0\right)$
• the distance between the foci is $\text{\hspace{0.17em}}2c,$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±c\right)$
• the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{a}{b}x$

Note that the vertices, co-vertices, and foci are related by the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}.\text{\hspace{0.17em}}$ When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
f(x)=x square-root 2 +2x+1 how to solve this value
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
24x^5
James
10x
Axmed
24X^5
Taieb
comment écrire les symboles de math par un clavier normal
SLIMANE