Question

1. (y - 7)^2 = -12(x + 1)

part i: what kind of conic section is this? (2 points)
part ii: what are the coordinates of the vertex of this graph? (2 points)
part iii: which direction does this shape open on the graph? (2 points)
part iv: what is the equation of the directrix? (2 points)
part v: what are the coordinates of the focus? (2 points)

Explanation:

Step1: Identify conic - section type

The given equation \((y - 7)^2=-12(x + 1)\) is in the form \((y - k)^2 = 4p(x - h)\), which represents a parabola.

Step2: Find vertex coordinates

For a parabola in the form \((y - k)^2=4p(x - h)\), the vertex is \((h,k)\). Comparing with \((y - 7)^2=-12(x + 1)\), we have \(h=-1\) and \(k = 7\). So the vertex is \((-1,7)\).

Step3: Determine opening direction

Since \(4p=-12\) (so \(p=- 3<0\)), for the form \((y - k)^2 = 4p(x - h)\), the parabola opens to the left.

Step4: Find directrix equation

For a parabola \((y - k)^2 = 4p(x - h)\), the directrix is given by the equation \(x=h - p\). Here \(h=-1\) and \(p=-3\), so \(x=-1-(-3)=2\). The directrix is \(x = 2\).

Step5: Find focus coordinates

For a parabola \((y - k)^2=4p(x - h)\), the focus is \((h + p,k)\). Substituting \(h=-1\), \(p=-3\) and \(k = 7\), we get \((-1-3,7)=(-4,7)\).

Answer:

Part I: Parabola
Part II: \((-1,7)\)
Part III: Opens to the left
Part IV: \(x = 2\)
Part V: \((-4,7)\)