advertisement

9.2 Graph & Write Equations of Parabolas Algebra II Parabolas • We already know the graph of y=ax2 is a parabola w/ vertex (0,0) and AOS 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 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 Focus (p,0) Directrix x=-p Standard equation of Parabola (vertex @ origin) Equation x2=4py y2=4px Focus (0,p) (p,0) Directrix AOS y=-p Vertical (x=0) x=-p Horizontal (y=0) Steps to graphing • 1.) isolate variable squared • 2.) Set coefficient of the variable to the first power equal to 4p; solve for p. • 3.) Identify focus & directrix (using table or pictures) • 4.) Make a T-chart accordingly. Ex. 1) Identify the focus and directrix of the parabola x = -1/6y2 EX. 1 • 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, x= -(3/2), p=3/2 • To draw: make a table of values & plot • p<0 so opens left so only choose neg Ex. 2 • Find the focus and directrix, then graph x= • • • • 2 3/4y 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 Standard equation of Parabola (vertex @ origin) Equation x2=4py y2=4px Focus (0,p) (p,0) Directrix AOS y=-p Vertical (x=0) x=-p Horizontal (y=0) Ex. 3) Writing the equation of a parabola. • The graph shows V=(0,0) • Directrex y=-p=-2 • So substitute 2 for p • = 4py 2 • x = 4(2)y • x2 = 8y 2 x • y = 1/8 and check in your calculator 2 x Standard equation of Parabola (vertex @ origin) Equation x2=4py y2=4px Focus (0,p) (p,0) Directrix AOS y=-p Vertical (x=0) x=-p Horizontal (y=0) Ex. 4 Write equation of parabola. • Focus = (0,-3) • X2 = 4py • X2 = 4(-3)y • X2 = -12y • y=-1/12x2 to check Assignment