Thread: parabolas - urgent

Hey guys, kinda stumped here on these 2. Especially the first one since it's a horizontal parabola, I don't really know how the form is supposed to be. Any help appreciated.

I know that standard form for a vertical is y = a(x - h)^2 + k, if that helps.

1. Write in standard form, showing work: -2y^2 + 4y + 6 = x

2. Fint the equation(in standard form) of the parabola that satisfies the following conditions: the vertex is at the origin, the parabola passes through the point (3,1), and the focus lies on the x-axis.

Again, any help is MUCH appreciated.

Originally Posted by leviathanwave

I know that standard form for a vertical is y = a(x - h)^2 + k, if that helps.

1. Write in standard form, showing work: -2y^2 + 4y + 6 = x

let's complete the square:

-2y^2 + 4y + 6 = x
=> -2(y^2 - 2y - 3) = x
=> -2(y^2 - 2y + (-1)^2 - (-1)^2 - 3) = x
=> -2((y - 1)^2 - 4) = x
=> -2(y - 1)^2 + 8 = x
=> x = -2(y - 1)^2 + 8

Originally Posted by leviathanwave

2. Fint the equation(in standard form) of the parabola that satisfies the following conditions: the vertex is at the origin, the parabola passes through the point (3,1), and the focus lies on the x-axis.

the equation of a parabola with focus (p,0) and directrix x = -p is

y^2 = 4px

since (3,1) lies on the parabola,

1^2 = 4p(3)
=> 1 = 12p
=> p = 1/12

so y^2 = (1/3)x
so x = 3y^2 is our parabola