***A projectile is launched at x=0 .y=h with speed V and angle θ. What is the equation of the parabola (no air drag)?**

y = ax2 + bx + c

a=(g/2)/(Vcosθ)2 … (g is negative)

b=tanθ

c=h

**Given 2 points (x[sub]1[/sub],y[sub]1[/sub]) and (x[sub]2[/sub],y[sub]2[/sub]), find the equation of the parabola that passes through the origin and those 2 points**

y = ax2 + bx + c

a=(x[sub]2[/sub]y[sub]1[/sub]-x[sub]1[/sub]y[sub]2[/sub])/(x[sub]1[/sub]2x[sub]2[/sub]-x[sub]1[/sub]x[sub]2[/sub]2)

b=-(x[sub]2[/sub]2y[sub]1[/sub]-x[sub]1[/sub]2y[sub]2[/sub])/(x[sub]1[/sub]2x[sub]2[/sub]-x[sub]1[/sub]x[sub]2[/sub]2)

c=0

**Given 2 points (x[sub]1[/sub],y[sub]1[/sub]) and (x[sub]2[/sub],y[sub]2[/sub]), find the equation of the parabola that passes through (0,h) and those 2 points**

y = ax2 + bx + c

a=(x[sub]1/sub+x[sub]2[/sub]y[sub]1[/sub]-hx[sub]2[/sub])/(x[sub]1[/sub]2x[sub]2[/sub]-x[sub]1[/sub]x[sub]2[/sub]2)

b=-(x[sub]1[/sub]2(h-y[sub]2[/sub])+x[sub]2[/sub]2y[sub]1[/sub]-hx[sub]2[/sub]2)/(x[sub]1[/sub]2x[sub]2[/sub]-x[sub]1[/sub]x[sub]2[/sub]2)

c=h

**Given 3 points (x[sub]1[/sub],y[sub]1[/sub]), (x[sub]2[/sub],y[sub]2[/sub]), and (x[sub]3[/sub],y[sub]3[/sub]) find the equation of the parabola that passes through those 3 points**

**Given parabola y = ax2 + bx + c , where a<0, find the value of x and y at the apex**

x = -b/(2a) … y=c-b2/(4a)

**Given parabola y = ax2 + bx + c , where a<0, find the value of x and y for which the slope is -1**

x = -(1+b)/(2a) … y = c+(b+1)2/(4a)-b(b+1)/(2a)

someone please check my math

parabola 3 points.xls (14.5 KB)

parabola 3 points.xls (14.5 KB)