Posted by Jody on Friday, April 25, 2008 at 5:29pm.
I'm sure how to work the following problems cant anyone help?
1) Write an equation for the parabola with focus (4,0) and directrix y=2.
What do I need to do before I graph the following equations?
e1) 16x^2+9y^2=144
e2) x^2/9  (y+2)^2/9 = 1

Algebra  Damon, Friday, April 25, 2008 at 5:52pm
the focus is at y = 0 and the directrix is at y = 2 so the parabola opens down (sheds water.)
The distance from the focus to the vertex is the same as from the directrix to the vertex ( call it a ) so the vertex is at (4,1), halfway between directrix and focus
now so a = 1 but if parabola is upside down, use negative a so a = 1
with vertex at (h,k) which is (4,1)
form is
(xh)^2 = 4 (a)(yk)
(x4)^2 = 4 (y1)
x^2  8x + 16 = 4 y + 4
 4 y = x^2  8 x + 16
y = (x^2)/4 + 2 x  4

Algebra  Damon, Friday, April 25, 2008 at 6:02pm
e1) 16x^2+9y^2=144
this is an ellipse
(16/144)x^2 + (9/144) y^2 = 1
x^2/3^2 + y^2/4^2 = 1
center at (0,0)
x half axis 3
y half axis 4
e2) x^2/3^2  (y+2)^2/3^2 = 1
hyperbola
center at (0, 2)
slopes of asymptotes = 3/3 = 1 (or 1)
vertex at (3,2) and at (3,2)
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