Question

What are the vertex, focus, and directrix of the parabola with the given equation? y=1/28(x-4)^2-5

Answer 1

Answer:

Vertex - (4,-5)

Focus (4,2)

Directrix y=-12

Step-by-step explanation:

The equation $y=\dfrac{1}{28}(x-4)^2-5$ of the parabola shows that its vertex is at point (4,-5). Multiply the equation by 28:

$28y=(x-4)^2-140\Rightarrow (x-4)^2=28y+140,\\ \\(x-4)^2=28(y+5).$

The parameter p of the parabola is

$2p=28\Rightarrow p=14.$

The coordinates of the focus will be

$\left(4,-5+\dfrac{p}{2}\right)=\left(4,-5+\dfrac{14}{2}\right)=\left(4,2\right).$

The directrix has the equation

$y=-5-\dfrac{p}{2}\Rightarrow y=-5-7,\ y=-12.$

Answer 2

The answer is:

vertex: (4,-5); focus: (4,2); directrix: y=-12