(x-3)^2=-12(y+1)

Question

I need to know how to graph this equation. What is the vertex, the directrix, focus, axis of symmetry and endpoints of the latus rectum.

Answer 1

y+1=(-1/12)(x-3)^2 is the form y-k=a(x-h)^2 in which the vertex (h,k) can be clearly seen as (3,-1). The line of symmetry is x=3, passing through the vertex (maximum) and focus, and the parabola is an inverted broad U shape, because a<0. The curve cuts the y axis (y intercept) when x=0, so y=-1-3/4=-7/4.

The focus is a distance 1/4a below the vertex at (3,-4) and the directrix line, y=2, is the same distance above the vertex at (3,2). The latus rectum is where the line y=-4 cuts the curve: -3=(-1/12)(x-3)^2; 36=(x-3)^2, x-3=+6, x=9, -3 so the end points are (-3,-4) and (9,-4).

 

Comments

Thank you so much this really helped. But how would I graph this? is there any way you show me how to graph this?

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Refer to the answer and mark on a graph all the points and lines given. The shape is a wide upturned U that should cut the y axis at y=-7/4 with its left arm. Remember the vertex is the highest point (maximum). You can draw the latus rectum as a line joining the points (-3,-4) and (9,-4) passing through the focus. To help you draw with reasonable accuracy, make a table of x-y values, starting with x=3 up to 9. The axis of symmetry is like a mirror so you can plot points by reflection on either side of the mirror. For example, y=-4 when x=9 and -3. For an exam you would only need to show the important points and the general shape.

Answer 2

Given (x-3)^2=-12(y+1)
Focus = (3,-4)
vertex = (3,-1)
semi-axis length = 3
focal parameter = 6

eccentricity = 1
directrix = y= 2

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