Find an equation of the parabola with vertex at the origin that passes through the point (-36, 2) and opens to the left.

I believe you start with (x-h)^2 = 4p(y-k) and plug in the given point and (0,0), and solve for p.

I got y^2 = 144x. Is this correct?

Thank you!

since the vertex is at the origin and the parabola opens to the left, it would have the simple form

x = -ay^2
plug in (-36,2) to get a = 9

so the equation is x = -9y^2

you should have realized your answer was wrong since the given point does not even satisfy your equation.
Secondly it would open to the right.

To find the equation of a parabola with the vertex at the origin and opening to the left, we can start with the standard form of a parabola equation: (x-h)^2 = 4p(y-k), where (h,k) is the vertex, and p is the distance from the vertex to the focus.

In this case, since the vertex is at the origin, we have h = 0 and k = 0. So our equation becomes x^2 = 4p(y-0), which simplifies to x^2 = 4py.

Next, we need to find the value of p. Given that the parabola passes through the point (-36, 2), we can substitute these coordinates into the equation: (-36)^2 = 4p(2). Simplifying this, we get 1296 = 8p. Dividing both sides by 8, we find p = 162.

So the equation of the parabola with the vertex at the origin, that passes through the point (-36, 2) and opens to the left, is x^2 = 4(162)y. Simplifying this further gives x^2 = 648y, or equivalently y = (1/648)x^2.

Therefore, the correct equation for the parabola is y = (1/648)x^2.