Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Write the equation of a hyperbola with vertices (0, -5) and (0, 5) and co-vertices (-2, 0) and (2, 0)

OpenStudy (asnaseer):

Here is a very good site that will explain how the various components in the equation of a hyperbola fit together: http://www.mathsisfun.com/geometry/hyperbola.html have a look through this and, if you still need help, then let me know and I will try and help you further.

OpenStudy (anonymous):

ok..ill try to work it out i guess

OpenStudy (asnaseer):

look also here: http://jwilson.coe.uga.edu/emt668/EMAT6680.2004.SU/Stein/6690/conic_sections_files/summary.htm the hyperbola is defined as the 3rd curve down. It contains all the components that your question mentions.

OpenStudy (anonymous):

ok so would i be looking for a horzontal or vertical graph

OpenStudy (asnaseer):

if you look carefully at the forms of the two types of hyperbola, then you should conclude that yours must be one with a Vertical Transverse Axis.

OpenStudy (anonymous):

ok, would it be (x-0)^2/2^2+(y-5)^2/0^2 ?

OpenStudy (asnaseer):

no - are you sure you solved for the vertical hyperbola?

OpenStudy (anonymous):

yes..well tried

OpenStudy (asnaseer):

its equation is given by:\[\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1\]and its vertices are at:\[(h,k-a),(h,k+a)\]and its co vertices are at:\[(h-b,k),(h+b,k)\]

OpenStudy (asnaseer):

look at the information given to you and let me know what you get for the values of a, b, h and k

OpenStudy (anonymous):

im confused on the a^2

OpenStudy (asnaseer):

first, look at the equations for the vertices of this hyperbola and form some relationships with the data given to you in your question

OpenStudy (asnaseer):

e.g. you are told that the vertices are at (0, -5) and (0, 5) therefore you can say that: (h,k-a) = (0,-5) (h,k+a) = (0, 5)

OpenStudy (asnaseer):

now compare each part of the coordinate pairs to establish some relationships

OpenStudy (asnaseer):

lets just take the first one: (h, k - a) = (0, -5) what does that tell you about h? what must it equal?

OpenStudy (anonymous):

h,k+a?

OpenStudy (asnaseer):

if (a, b) = (c, d) then it implies a = c and b = d so, if (h, k - a) = (0, -5) then h = ?

OpenStudy (anonymous):

0

OpenStudy (asnaseer):

correct, now use the same techniques to find the other values (k, a and b)

OpenStudy (anonymous):

(y-0)^2?

OpenStudy (asnaseer):

please list the values you have found for h, k, a and b

OpenStudy (anonymous):

h=0, k - a) = (0, -5)

OpenStudy (asnaseer):

you need to use all the information given to you to find values for h, k, a and b. I showed you how to do the first part where I equated the equations for the vertices to the coordinates of the vertices given to you in the question. you need to do the same now with the co vertices.

OpenStudy (anonymous):

(h−b,k),(h+b,k) = -2,0 2,0

OpenStudy (asnaseer):

can you please explain what you are doing?

OpenStudy (asnaseer):

I want to see if I can spot where the confusion lies

OpenStudy (asnaseer):

?

OpenStudy (anonymous):

Im writing the co vertices

OpenStudy (asnaseer):

yes, but I asked you to use these to solve for h, k, a and b - but instead you just list the relations

OpenStudy (asnaseer):

lets try from the beginning - step by step...

OpenStudy (asnaseer):

what are the coordinates of the vertices that have been given to you in this question?

OpenStudy (anonymous):

(0,-5) and (0,5)

OpenStudy (asnaseer):

good. now what are the coordinate of the vertices of a vertical hyperbola in terms of h, k, a and b?

OpenStudy (asnaseer):

(I listed these for you up there ^^^)

OpenStudy (anonymous):

(-2,0)?

OpenStudy (asnaseer):

no, the vertices will be at: (h,k-a) and (h,k+a)

OpenStudy (asnaseer):

It is very late where I am at so I need to go now - maybe someone else can try to explain this to you.

OpenStudy (anonymous):

can you not just give me the answer before you get off? im sorry im reallly slow...but i have to have this problem done :/

OpenStudy (asnaseer):

sorry - the aim of this site is to teach - not just give out answers. it doesn't matter if you are slow - you need to persevere and try to understand the concepts here otherwise you will never understand how to solve these. anyway - good luck...

OpenStudy (anonymous):

k bye.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!