Question

the focus, (0, -2), and directrix, x = 6, of a parabola are graphed below. which equation represents the parabola? x = 1/12(y + 2)^2+3 x=-1/12(y + 2)^2+3 x=-1/12(y - 2)^2-3 x = 1/12(y - 2)^2-3

Explanation:

Step1: Recall parabola definition

The distance from any point \((x,y)\) on the parabola to the focus \((x_f,y_f)=(0, - 2)\) is equal to the distance from the point \((x,y)\) to the directrix \(x = 6\). The distance formula from \((x,y)\) to the focus is \(d_1=\sqrt{(x - x_f)^2+(y - y_f)^2}=\sqrt{(x - 0)^2+(y + 2)^2}\), and the distance from \((x,y)\) to the directrix \(x = 6\) is \(d_2=\vert x - 6\vert\).

Step2: Set up the equation

\(\sqrt{x^{2}+(y + 2)^{2}}=\vert x - 6\vert\). Square both sides: \(x^{2}+(y + 2)^{2}=(x - 6)^{2}\). Expand \((x - 6)^{2}=x^{2}-12x + 36\). So \(x^{2}+(y + 2)^{2}=x^{2}-12x + 36\).

Step3: Simplify the equation

Cancel out \(x^{2}\) on both sides: \((y + 2)^{2}=-12x + 36\). Rearrange to solve for \(x\): \(12x=- (y + 2)^{2}+36\), then \(x=-\frac{1}{12}(y + 2)^{2}+3\).

Answer:

\(x =-\frac{1}{12}(y + 2)^{2}+3\)