Please explain how to find p, the focal point, step-by-step, in the following equation: -3y^2 -12x=0. Thanks!

Answer 1

The standard form of parabola is y² = - 4ax with vertex at (0,0) and focus at (-a, 0).

Our function is -3y² - 12x = 0 so let's rearrange to conform:

=> 3y² = -12x

=> y² = -4x

hence a = 1 and our vertex is (0,0) and the focus is at (-1, 0)

Answer 2

The standard form of a parabola is

(x-h)^2 = 4p(y - k); the vertex is at (h, k), and the focus is at (h, k + p)

With this problem, -3y^2 - 12x = 0

-3y^2 = 12x

y^2 = -4x

Without doing any work, except for the minus sign, we are already in the standard form, except y and x are switched in the equation.

Thus, p = -1.(-1 * 4 = -4, as found in the equation).

the vertex is at the origin, (0, 0) - h and k are 0.

Thus, the focus is at (k + p, h) - switching x and y - so the focus is at

(0 + -1, 0) = (-1, 0)