Hey gues find equation of parabola with given properties. Also a hyperbola question.

**dham** · April 29th 2009, 07:36 PM

Hey guys I have a question I've been stuck on. The problem says

Find the equation of the parabola with the given properties.
Vertex(1,3); vertical axis of symmetry; containing the point(-1,19)

I can't find this in my book anywhere.

I also have on that says Find the center, foci, and asymptotes of the hyperbola.

7x^2 - y^2 - 42x + 8y - 65 = 0

I think I have the completing the square down. I'm not sure though
I get stuck at this part

7(x^2 - 6x) - 1(y^2 - 8y) = -65

I'm stuck what to do with the completing the square after that. After that I'm not sure on how to find the center, asymptotes and foci.

And I hope I'm not asking to much. But I have this question to. I'm not sure where to start.

Find the equation of the hyperbola satisfying the given conditions.
Vertices: (-6,4) and (8,4); asymptotes y = 2x + 8 and y = -2x + 0

Thanks for any help, steps, or answers guys. I appreciate a lot.

**pickslides** · April 29th 2009, 09:44 PM

Originally Posted by dham

Hey guys I have a question I've been stuck on. The problem says

Find the equation of the parabola with the given properties.
Vertex(1,3); vertical axis of symmetry; containing the point(-1,19)

I would use the equation for turning point

$y=a(x-h)^2+k$

Where a is the dialation (determines the steepness of the function), h is the horizontal translation and k is the vertical translation.

If the vertex is at (1,3) then we can say

$y=a(x-1)^2+3$ now substitute the point (-1,19) into this equation to find a.

$19=a(-1-1)^2+3$

$19=a(-2)^2+3$

$19=4a+3$

$16=4a$

$a=4$

therefore the equation for this parabola is,

$y=4(x-1)^2+3$

**pflo** · April 30th 2009, 01:44 PM

Originally Posted by dham

I think I have the completing the square down. I'm not sure though.
I get stuck at this part: $7(x^2 - 6x) - 1(y^2 - 8y) = -65$

I'm stuck what to do with the completing the square after that.
After that I'm not sure on how to find the center, asymptotes and foci.

You would have $7[(x-3)^2-9]-1[(y-4)^2-16]=-65$ after completing the square.

Rearrange to put this into the standard form for a hyperbola: $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ (standard form when the 'y-squared' term is negative, which results in a horizontal transverse axis - the equations below change slightly if the hyperbola has a vertical transverse axis).

The center of the hyperbola is at $(h,k)$ and both asymptotes go through this point. The slopes of the asymptotes are ${\pm}\frac{b}{a}$
The vertices are at $({\pm}a+h,k))$
The foci are at $({\pm}c+h,k)$ where $c=\sqrt{a^2+b^2}$

**stapel** · April 30th 2009, 01:59 PM

Originally Posted by dham

I can't find this in my book anywhere.

Ouch!

To replace the missing lessons, try studying from some online resources. Click for explanations of parabolas and hyperbolas. If you're also needing help with the other conics, click for explanations of circles and ellipses.

Good luck!

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