parabola

JoanneMac · #1 $Old$ 12-03-2008, 01:41 PM

that are busting my chops today. Hopefully I can get them both solved.

1. Find the directrix, focus, and the roots of the parabola y=x^2-5x+4.

2. Graph x^2/16-y^2/25=1. Show how you arrived at your graph by determining the x-intercepts, extent of the graph, and the asymptotes.

I figured out the x-intercepts as being 4 and -4, but I have no clue as to what they mean by extent of the graph and asymptotes don't ring a bell with me.

Anyone's help is greatly appreciated.

Joanne

skeeter · #2 $Old$ 12-03-2008, 05:44 PM

$y = x^2 - 5x + 4$

roots ...

$x^2 - 5x + 4 = 0$

$(x - 1)(x - 4) = 0$

see the roots?

to get the focus and directrix, need to find the vertex coordinates ...

$x = \frac{-b}{2a} = \frac{5}{2}$

$y = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 4$

$y = -\frac{9}{4}$

focal length, $d = \frac{1}{4a} = \frac{1}{4}$

focus is at $\left(\frac{5}{2} , -\frac{9}{4} + \frac{1}{4}\right) = \left(\frac{5}{2} , -2\right)$

directrix is $y = -\frac{9}{4} - \frac{1}{4} = -\frac{5}{2}$

general form for a hyperbola centered at the origin ...

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

asymptotes are $y = \frac{b}{a} x$ and $y = -\frac{b}{a} x$

focal length $c = \sqrt{a^2 + b^2}$ ... focal points are $(-c,0)$ and $(c,0)$