Hyperbola: Find Vertices, Foci & Asymptotes

[js14]

Find the vertices, foci, and asymptotes of the hyperbola.
x^2 − 7y^2 = 8
When I tried to solve this i got + or - 2 sqrt 2 for the vertices, + or - sqrt 64/7,0 as the focus and + or minus sqrt 65/7 divided by 2 sqrt 2x as the asymptotes. is this correct?

 

[eumyang]

Find the vertices, foci, and asymptotes of the hyperbola.
x^2 − 7y^2 = 8
When I tried to solve this i got + or - 2 sqrt 2 for the vertices

I would consider these not completely correct because these are not expressed as points. Can you tell us what the points are exactly?

+ or - sqrt 64/7,0 as the focus... (snip)

This is a little better, because at least you mention the 0. But where are the parentheses? Also, do you mean
$$\pm\sqrt{\frac{64}{7}}$$
or
$$\pm\frac{\sqrt{64}}{7}$$
? It is not clear to me from what you wrote.

 

[js14]

When I tried to solve this i got + or - 2 sqrt 2,0 for the vertices, + or - sqrt (64/7),0 as the focus and + or minus sqrt (65/7) divided by 2 sqrt 2x as the asymptotes. do you understand this better?

 

[eumyang]

When I tried to solve this i got + or - 2 sqrt 2,0 for the vertices, + or - sqrt (64/7),0 as the focus and + or minus sqrt (65/7) divided by 2 sqrt 2x as the asymptotes. do you understand this better?

Points are written like this:
(5, -9)
(-2, 6)

For your vertices:

+ or - 2 sqrt 2,0

are still not written as points. At least they aren't to me. I'm sorry to be picky, but unless you are not an English speaker and this is the norm for you, this comes across as sloppy to me.

For your asymptotes:

+ or minus sqrt (65/7) divided by 2 sqrt 2x

Oh, you mean this?
$$\pm \frac{\sqrt{65/7}}{2\sqrt{2x}}$$

I know you don't mean that, but again, the way you typed that is unclear.

 

[js14]

Ok the the points for the vertices are (2 sqrt 2,0) (-2 sqrt 2,0) for the foci I have points (sqrt (64/7),0) (-sqrt(-64/7),0) and the same radical that you put in your last post + or minus sqrt (65/7) divided by 2 sqrt 2 x as the asymptotes . except the x shoulnd be in the denomonator it shound be to the right of the radical. Now i that I have the vertices and foci right because its in my online homework and it says I am right. where I am confused is with the asymptotes. if i divide b/a then i get the radical (65/7) divided by 2 sqrt 2. if I simplifiy that i get 1/2 * sqrt(65/14). I thought that this was the answer but my it apparently its not because it is marked as wrong. What I am I doing wrong?

 

[HallsofIvy]

Ok the the points for the vertices are (2 sqrt 2,0) (-2 sqrt 2,0) for the foci I have points (sqrt (64/7),0) (-sqrt(-64/7),0) and the same radical that you put in your last post + or minus sqrt (65/7) divided by 2 sqrt 2 x as the asymptotes . except the x shoulnd be in the denomonator it shound be to the right of the radical.

Nonsense! The asymptotes are straight lines and it would make no sense to put x in a denominator. Here is the easy way to get the asymptotes: The equation of your hyperbola is $x^2- 7y^2= 8$. For very, very, very large x and y the two terms on the left will be so much larger than the "8" on the right that we can "ignore" it. That is, the graph of the hyperbola will be very close to the graph of $x^2-7y^2= (x- \sqrt{7}y)(x+ \sqrt{7}y)= 0$. That is, the two lines $x= \sqrt{y}$ (or $y= x/\sqrt{7}$ and $x= -\sqrt{7}y$ (or $y= -x/\sqrt{7}$)

Now i that I have the vertices and foci right because its in my online homework and it says I am right. where I am confused is with the asymptotes. if i divide b/a then i get the radical (65/7) divided by 2 sqrt 2. if I simplifiy that i get 1/2 * sqrt(65/14). I thought that this was the answer but my it apparently its not because it is marked as wrong. What I am I doing wrong?

 

[eumyang]

Ok the the points for the vertices are (2 sqrt 2,0) (-2 sqrt 2,0) for the foci I have points (sqrt (64/7),0) (-sqrt(-64/7),0) and the same radical that you put in your last post + or minus sqrt (65/7) divided by 2 sqrt 2 x as the asymptotes . except the x shoulnd be in the denomonator it shound be to the right of the radical. Now i that I have the vertices and foci right because its in my online homework and it says I am right. where I am confused is with the asymptotes. if i divide b/a then i get the radical (65/7) divided by 2 sqrt 2. if I simplifiy that i get 1/2 * sqrt(65/14). I thought that this was the answer but my it apparently its not because it is marked as wrong. What I am I doing wrong?

As HallsOfIvy said, x is not in the denominator. Also, I'm not getting
$$b = \sqrt{\frac{65}{7}}$$
so it looks like that's part of the problem. What does the equation look like when you wrote it in the form of
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
?

 

[js14]

i have x^2/8 + y^2/(8/7) =1.

 

[vela]

Just out of curiosity, how did you get the 65 that appeared under the radical?

 

[BloodyFrozen]

Find the vertices, foci, and asymptotes of the hyperbola.
x^2 − 7y^2 = 8
When I tried to solve this i got + or - 2 sqrt 2 for the vertices, + or - sqrt 64/7,0 as the focus and + or minus sqrt 65/7 divided by 2 sqrt 2x as the asymptotes. is this correct?

I'm not sure if I'm right or coooooooooooooooooooommmmmmmmmmppppppplllllleeeeeeteeeeeelllllly wrong, but...

What I'd do:

First simplify---

x^2/8 - 7y^2/8 =1

Then...

x^2/8 - y^2/(8/7)=1
You see how I got this?

This hyperbola has a horizontal transverse axis, so...

Vertices: (sqrt(8),0) ; (-sqrt(8),0) <---------------Simplify the radical if you want
Covertices: (0, sqrt(8/7)) ; (0,-sqrt(8)) <------------Again, Simplify if you want

Remember foci has to do with c

a^2+b^2= c^2
8 + (8/7) =c^2
64/7= c^2 <<<<<<<<<<<<<<<Simplify
(8*sqrt(7))/7 = c

So.. the foci is

[(8*sqrt(7))/7,0] and [-(8*sqrt(7))/7,0]

How I remember asymptotes for hyperbola is: +or- number under y^2 divided by number under x^2, but make sure it's raised to ^2

So for your example,

Asymptotes: y= [(8/7)/8]x or y= -[(8/7)/8]x <<<<This will yield...
y= (1/7)x or y= (-1/7)x

Atleast that's how I do it... Correct me if I'm wrong