Expert: Abe Mantell - 7/16/2009

Question
1.State the vertex of  x2 – 2x – y = 0.
2.State the equation of the directrix of   (y – 4)2 =  –8(x – 3) .
3.State the equation of the axis of symmetry of  x2 – 6x + 10y – 1 = 0.
4.State the direction of the opening of  y2 + 8x – 6y – 3 = 0.
5.State the coordinates of the focus of  x2 – 4y – 8 = 0
6. Write the general equation of the parabola given vertex (4, 3)  and focus (6, 3).


Answer
1. y=x^2-2x has vertex at x=-b/(2a)=2/2=1, so x=1 and y=-1 ==> (1,-1)
2. Basically, if you arrange the equation for a parabola in this form:
(x-h)^2=4p(y-k)
The vertex is (h,k); the focus is (h, k+p); the directrix is y=k-p where p is the distance from the vertex to the focus.
or if of the form (y-h)^2=4p(x-k) has directrix x=k-p
You have (y-4)^2=-8(x-3) ==> (y-4)^2=4(-2)(x-3)...
so the directrix is x=3-(-2) or x=5
3. x^2-6x-1=-10y ==> y=(-1/10)x^2+(3/5)x+(1/10)
.. axis of symmetry is x=-b/(2a)=-(3/5)/(-2/10) .. x=3
4. y^2-6y-3=-8x or 8x=-y^2+6y+3 since the y^2 is negative it "opens"
.. to the left.
5. 4y=x^2-8 can be written as x^2=4y+8 or (x-0)^2=4(1)(y+4)
.. so, as explained in #2, the focus is (0,-4+1) = (0,-3)
6. (y-3)^2=4(2)(x-4), the 2 comes from 6-4...

Check this site for lots of useful info about parabolas:
http://en.wikipedia.org/wiki/Parabola

Abe