Finding the equation of a hyperbola given certain conditions

In summary, the homework statement is trying to find the equation of a hyperbola whose transverse axis is x = 3 and goes through the center of the circle.
  • #1
supermiedos
63
0

Homework Statement


Find the equation of the hyperbola whose transverse axis is x = 3 and goes through:

The vertices of 2x^2 + y^2 - 28x + 8y + 108 = 0 and the center of
x^2 + y^2 - 6x + 4y + 3 = 0.

Homework Equations


(x - h)^2/a^2 - (y - k)^2/b^2 = 1

The Attempt at a Solution


so far I have identified the vertices of the ellipse and the center of the circumference, as you can see here:

http://i.imgur.com/QleqyMH.png

But now I don't have any idea how to proceed next. Could you help me please?
 
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  • #2
hi supermiedos! :smile:
supermiedos said:
Find the equation of the hyperbola whose transverse axis is x = 3 and goes through:

… the center of
x^2 + y^2 - 6x + 4y + 3 = 0.

so far I have identified the center of the circumference …

nooo, it's the centre of the circle :wink:

(how can a circumference have a centre? :confused:)
 
  • #3
supermiedos said:

Homework Equations


(x - h)^2/a^2 - (y - k)^2/b^2 = 1
A little nitpick, but this is the wrong equation. The equation you wrote requires a horizontal transverse axis.

supermiedos said:

The Attempt at a Solution


so far I have identified the vertices of the ellipse and the center of the circumference, as you can see here:

http://i.imgur.com/QleqyMH.png

But now I don't have any idea how to proceed next. Could you help me please?
If the center of the circle is a point on the hyperbola, then surely this point is also one of the two vertices of the hyperbola, is it not? See if you can take it from there.
 
Last edited:
  • #4
tiny-tim said:
hi supermiedos! :smile:


nooo, it's the centre of the circle :wink:

(how can a circumference have a centre? :confused:)

:tongue: you are right, my bad
 
  • #5
eumyang said:
A little nitpick, but this is the wrong equation. The equation you wrote requires a horizontal transverse axis.


If the center of the circle is a point on the hyperbola, then surely this point is also one of the two vertices of the hyperbola, is it not? See if you can take it from there.

Thank you for your help, I finally did it. Thank you so much
 

1. How do I find the equation of a hyperbola when given the center and vertices?

In order to find the equation of a hyperbola, you will need to use the standard form equation: (x - h)^2/a^2 - (y - k)^2/b^2 = 1, where (h, k) is the center of the hyperbola and a and b are the distances from the center to the vertices along the x and y axes, respectively.

2. Can I find the equation of a hyperbola if I am only given the foci and a point on the curve?

Yes, you can still find the equation of a hyperbola with this information. You will need to use the distance formula to find the distances from the foci to the given point, and then use those distances to solve for a and b in the standard form equation.

3. Is it possible to find the equation of a hyperbola if I am only given the asymptotes?

No, you will need more information in order to find the equation of a hyperbola. The asymptotes alone do not provide enough information to determine the center, vertices, or foci of the hyperbola.

4. How can I find the equation of a hyperbola if I am given the eccentricity and the distance between the foci?

If you are given the eccentricity (e) and the distance between the foci (2c), you can use the formula c^2 = a^2 - b^2 to solve for a and b. Then, you can use these values to find the equation of the hyperbola using the standard form equation.

5. Can I find the equation of a hyperbola if I am given the asymptotes and one focus?

Yes, with this information, you can find the equation of the hyperbola. First, use the asymptotes to determine the center of the hyperbola. Then, use the distance formula to find the distance from the given focus to the center, and use this distance to solve for a in the standard form equation. Finally, use the distance formula again to find the distance from the center to the other focus, and use this distance to solve for b in the standard form equation.

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