[Ether]

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 (x₁,y₁) and (x₂,y₂), find the equation of the parabola that passes through the origin and those 2 points

y = ax² + bx + c

a=(x₂y₁-x₁y₂)/(x₁²x₂-x₁x₂²)

b=-(x₂²y₁-x₁²y₂)/(x₁²x₂-x₁x₂²)

c=0


Given 2 points (x₁,y₁) and (x₂,y₂), find the equation of the parabola that passes through (0,h) and those 2 points

y = ax² + bx + c

a=(x₁(h-y₂)+x₂y₁-hx₂)/(x₁²x₂-x₁x₂²)

b=-(x₁²(h-y₂)+x₂²y₁-hx₂²)/(x₁²x₂-x₁x₂²)

c=h


Given 3 points (x₁,y₁), (x₂,y₂), and (x₃,y₃) find the equation of the parabola that passes through those 3 points

It gets messy.


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