Finding the equation to hyperbola

[Niaboc67]

Homework Statement

[/B]

Homework Equations

General Hyperbola form: x^2/a^2 - y^2/b^2 = 1 or y^2/a^2 - x^2/b^2 = 1[/B]

The Attempt at a Solution

I am confused by this because I think I am close to the question but I have something fundamentally wrong.

What I know is since it's in a horizontal form it's going to be x^2/a^2 - y^2/b^2

Then I know that the vertex is (-6,0) so then the vertices is (-/+6,0) or rather √(36) so would this be x^2/36 - y^2/b^2 at this point? I don't see a way to get b^2 after this?

Another conflict I have here is that I know the asymptotes are y= -/+ b/a for this. So for b to become 1 it b must be √(1) and for a to become 5 it must be √(25) so then I come to the conclusion that the equation must be:

x^2/1 - y^2/√(25) = 1, but that wouldn't make any sense since the higher denominator is what makes it vertical/horizontal and since it's in the horizontal form the x^2 must have a higher denominator.

Please help
Thank you

 

[jedishrfu]

Does it help if you know that 0,6 and 0,-6 are points on the curve?

 

[Niaboc67]

What do you mean by 'the curve'?

 

[jedishrfu]

Isn't the red line on your graph the hyperbolic curve?

 

[Niaboc67]

Yeah, but wouldn't it be the points (-6,0),(6,0) since we are dealing with the x-axis?

 

[SteamKing]

Homework Equations

General Hyperbola form: x^2/a^2 - y^2/b^2 = 1 or y^2/a^2 - x^2/b^2 = 1[/B]

The Attempt at a Solution

I am confused by this because I think I am close to the question but I have something fundamentally wrong.

What I know is since it's in a horizontal form it's going to be x^2/a^2 - y^2/b^2

Then I know that the vertex is (-6,0) so then the vertices is (-/+6,0) or rather √(36) so would this be x^2/36 - y^2/b^2 at this point? I don't see a way to get b^2 after this?

Another conflict I have here is that I know the asymptotes are y= -/+ b/a for this. So for b to become 1 it b must be √(1) and for a to become 5 it must be √(25) so then I come to the conclusion that the equation must be:

x^2/1 - y^2/√(25) = 1, but that wouldn't make any sense since the higher denominator is what makes it vertical/horizontal and since it's in the horizontal form the x^2 must have a higher denominator.

Please help
Thank you

The slope of the asymptotes is ±b/a. You know the location of the vertices of the hyperbola and you are given the slopes of the asymptotes, both of which can be determined from the graph.

This article gives a handy reference to dissecting the various conic sections for their key properties:

http://en.wikipedia.org/wiki/Conic_section

 

[jedishrfu]

Yeah, but wouldn't it be the points (-6,0),(6,0) since we are dealing with the x-axis?

Yes you're right. I'm dislexic sometimes. Given that and the line slopes it seems you have what you need to find the equation.

SteamKing has provided a good reference too that will help you over any difficulties.

 

[Niaboc67]

That does help. But could someone point out what I am misunderstand or where I am going wrong with this?

 

[jedishrfu]

plugging the points in gives you the a value and using the asymtopes line equation gives you the b value right?

 

[SteamKing]

That does help. But could someone point out what I am misunderstand or where I am going wrong with this?

Another conflict I have here is that I know the asymptotes are y= -/+ b/a for this. So for b to become 1 ...

Why must b become 1?

You know that b/a = 1/5, and you know what a is from the graph. Algebra tells you b = a * 1/5 = a / 5.

 

[Niaboc67]

I thought since it states in my book y= -/+ b/a for a vertical hyperbola then it's the same as saying y = √b / √a ? I know that the sqrt of 1 is 1 and the sqrt of 25 is 5.

 

[jedishrfu]

Okay well 4/9 is not the same as 2/3 even though I've taken the square root of both numbers

 

[SteamKing]

I thought since it states in my book y= -/+ b/a for a vertical hyperbola then it's the same as saying y = √b / √a ? I know that the sqrt of 1 is 1 and the sqrt of 25 is 5.

You are assuming that just because b/a = 1/5 that a = 5 and b = 1.

The vertex of this hyperbola is not located at (5,0) and (-5,0); check the graph carefully. a must be equal to 6.

 

[Niaboc67]

Then shouldn't the asymptotes be y=√(36)/√(25) but then that would result in the asymptote sbeing y=±6/5

 

[Niaboc67]

I think I see where I am getting confused. I thought that since my book defines the vertices are as (±a,0) for the horizontal type then whatever 'a' is under the x^2 must be the same a when referring to the asymtopes.

 

[SteamKing]

I think I see where I am getting confused. I thought that since my book defines the vertices are as (±a,0) for the horizontal type then whatever 'a' is under the x^2 must be the same a when referring to the asymtopes.

But the a's are the same. It's the value of b which was getting you confused.

Since a = 6 and b / a = 1 / 5, then b / 6 = 1 / 5, which means b = ?

 

[Niaboc67]

b/6 = 1/5 cross-multiply and 5b/5 = 6/5 = 1.2?

 

[SteamKing]

b/6 = 1/5 cross-multiply and 5b/5 = 6/5 = 1.2?

Well, just b = 6 / 5. No need to cross multiply, just standard algebra clean up.

You can keep the fractional form and substitute back into the standard equation for the hyperbola.

x² / 6² - y² / (6/5)² = 1

x² / 36 - 25y² / 36 = 1

x² - 25y² = 36

 

[Niaboc67]

I am confused about two things here. Why does -y^2/(6/5)^2 turn into 25y^2/36? if it's in the denominator wouldn't it just be y^2/(36/25)?
And if the asymptotes is y=-/+1/5 and the final form is x^2/36 - 25y^2/36 = 1 wouldn't the asymptote be c=√(36+36) or 6√(2)?

 

[SteamKing]

I am confused about two things here. Why does -y^2/(6/5)^2 turn into 25y^2/36? if it's in the denominator wouldn't it just be y^2/(36/25)?

1 / (a / b)² = b² / a²

And if the asymptotes is y=-/+1/5 and the final form is x^2/36 - 25y^2/36 = 1 wouldn't the asymptote be c=√(36+36) or 6√(2)?

See above. The standard form of the equation of the hyperbola is x² / a² + y² / b² = 1

a² = 36, ∴ a = 6

b² = 36 / 25, ∴ b = 6 / 5

The slope of the asymptotes m = ± b / a = ± (6 / 5) / 6 = ± 1 / 5

 

[Niaboc67]

I understand now.