Locus and the Parabola Question (1 Viewer)

[Aron]

Challenge question 7

Find the equation of the parabola with axis parallel to the y-axis and passing through points (0,-2), (1,0) and (3,-8)

Thanks in advance!

 

[Study notes]

I got (x-1)^2 = -4(1/8)y

or

y = -2x^2 + 4x - 2 (expanded from the top)


idk if this is right, if it is... then i'll show working out.

Basically, from inspection I got a concave down parabola which has an equation of (x-k)^2 = -4a(y-h) which is just the general equation of a concave down parabola where k and h represents the vertex of the parabola at (k , h)

then sub the 3 points into that equation to simultaneously solve for k, h and a. Then put it back into that general equation

 

[obliviousninja]

I got (x-1)^2 = -4(1/8)y

or

y = -2x^2 + 4x - 2 (expanded from the top)


idk if this is right, if it is... then i'll show working out.

Basically, from inspection I got a concave down parabola which has an equation of (x-k)^2 = -4a(y-h) which is just the general equation of a concave down parabola where k and h represents the vertex of the parabola at (k , h)

then sub the 3 points into that equation to simultaneously solve for k, h and a. Then put it back into that general equation

Thats pretty much the method...so should be right if you haven't made a silly.

 

[Drongoski]

I got (x-1)^2 = -4(1/8)y

I believe that's correct.