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

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.

Completing the Square & Hyperbolas

Quote:

Originally Posted by

**dham**

I think I have the completing the square down. I'm not sure though.

I get stuck at this part: $\displaystyle 7(x^2 - 6x) - 1(y^2 - 8y) = -65$

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

Rearrange to put this into the standard form for a hyperbola: $\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ (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 $\displaystyle (h,k)$ and both asymptotes go through this point. The slopes of the asymptotes are $\displaystyle {\pm}\frac{b}{a}$

The vertices are at $\displaystyle ({\pm}a+h,k))$

The foci are at $\displaystyle ({\pm}c+h,k)$ where $\displaystyle c=\sqrt{a^2+b^2}$