Finding the equation of a hyperbola given certain conditions

[supermiedos]

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?

 

[tiny-tim]

hi supermiedos!

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

(how can a circumference have a centre? )

 

[eumyang]

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.

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.

 

[supermiedos]

hi supermiedos!


nooo, it's the centre of the circle

(how can a circumference have a centre? )

you are right, my bad

 

[supermiedos]

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