Question

1. (20 pts) derive the equation of the parabola with a focus at (-2, 4) and a directrix of y = 6. put the equation in standard form. (2 points) (a) (x + 2)^2=-1/4(y - 5) (b) (x + 2)^2=-1/4(y - 4) (c) (x + 2)^2=1/4(y + 5) (d) (x + 2)^2=1/4(y + 4)

Explanation:

Step1: Recall the definition of a parabola

The distance from a point \((x,y)\) on the parabola to the focus \((- 2,4)\) is equal to the distance from the point \((x,y)\) to the directrix \(y = 6\). The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), and the distance from a point \((x,y)\) to the horizontal line \(y = k\) is \(|y - k|\). So, \(\sqrt{(x + 2)^2+(y - 4)^2}=|y - 6|\).

Step2: Square both sides

\((x + 2)^2+(y - 4)^2=(y - 6)^2\). Expand the expressions: \(x^{2}+4x + 4+y^{2}-8y + 16=y^{2}-12y + 36\).

Step3: Simplify the equation

Cancel out \(y^{2}\) on both sides: \(x^{2}+4x + 4-8y + 16=-12y + 36\). Move the \(y\) - terms to one - side: \(12y-8y=-x^{2}-4x + 36 - 4 - 16\).

Step4: Combine like - terms

\(4y=-x^{2}-4x + 16\). Then \(y=-\frac{1}{4}x^{2}-x + 4\).

Answer:

The equation of the parabola is \(y =-\frac{1}{4}x^{2}-x + 4\)