Question

question 4 (2 points)
(02.08)
derive the equation of the parabola with a focus at (2, -1) and a directrix of y = -\frac{1}{2}.

a
f(x)=-(x + 2)^2-\frac{3}{4}

b
f(x)=(x + 2)^2+\frac{3}{4}

c
f(x)=-(x - 2)^2+\frac{3}{4}

d
f(x)=-(x - 2)^2-\frac{3}{4}

Explanation:

Step1: Recall the distance formula

The distance between a point $(x,y)$ on the parabola and the focus $(2,-1)$ is $\sqrt{(x - 2)^2+(y + 1)^2}$. The distance between the point $(x,y)$ and the directrix $y=-\frac{1}{2}$ is $\vert y+\frac{1}{2}\vert$.

Step2: Set the distances equal

Since a point on a parabola is equidistant from the focus and the directrix, we have $\sqrt{(x - 2)^2+(y + 1)^2}=\vert y+\frac{1}{2}\vert$. Square both sides: $(x - 2)^2+(y + 1)^2=(y+\frac{1}{2})^2$.

Step3: Expand the equations

Expand the right - hand side: $(x - 2)^2+y^{2}+2y + 1=y^{2}+y+\frac{1}{4}$.

Step4: Simplify the equation

Cancel out $y^{2}$ on both sides: $(x - 2)^2+2y + 1=y+\frac{1}{4}$. Then, move the terms involving $y$ to one side: $(x - 2)^2=y+\frac{1}{4}-2y - 1$. Combine like terms: $(x - 2)^2=-y-\frac{3}{4}$. Solve for $y$: $y=-(x - 2)^2-\frac{3}{4}$.

Answer:

D. $f(x)=-(x - 2)^2-\frac{3}{4}$