9.2 Graph and Write Equations of Parabolas

Parabolas We already know the graph of y=ax2 is a parabola w/ vertex (0,0) and AOS (axis of symmetry) x=0 Every parabola has the property that any point on it is equidistant from a point called the Focus and a line called the directrix.

Focus Lies on AOS Directrix

The focus and directrix each lie IpI units from the vertex The focus and directrix each lie IpI units from the vertex. (the vertex is ½ way between the focus and directrix)

x2=4py, p>0 Focus (0,p) Directrix y=-p

x2=4py, p<0 Directrix y=-p Focus (0,p)

y2=4px, p>0 Directrix x=-p Focus (p,0)

y2=4px, p<0 Directrix x=-p Focus (p,0)

Standard equation of Parabola (vertex @ origin) Focus Directrix AOS x2=4py (0,p) y=-p Vertical (x=0) y2=4px (p,0) x=-p Horizontal (y=0)

Identify the focus and directrix of the parabola x = -1/6y2 Since y is squared, AOS is horizontal Isolate the y2 → y2 = -6x Since 4p = -6 p = -6/4 = -3/2 Focus : (-3/2,0) Directrix : x=-p=3/2 To draw: make a table of values & plot p<0 so opens left so only choose neg values for x

Your Turn! y2 so AOS is Horizontal Isolate y2 → y2 = 4/3 x Find the focus and directrix, then graph x = 3/4y2 y2 so AOS is Horizontal Isolate y2 → y2 = 4/3 x 4p = 4/3 p = 1/3 Focus (1/3,0) Directrix x=-p=-1/3

Writing the equation of a parabola. The graph shows V=(0,0) Directrex y=-p=-2 So substitute 2 for p

x2 = 4py x2 = 4(2)y x2 = 8y y = 1/8 x2 and check in your calculator

Your turn! Focus = (0,-3) X2 = 4py X2 = 4(-3)y X2 = -12y y=-1/12x2 to check