SOLUTION: Please hepl solve this problem, vertex (2, 1), ends of the latus rectum (-1, -5) and (-1,7). (y-k)^2= -4c(x-h) (y- 1)^2= -4*3(x- 2) y^2-2y+1= 12(x-2)

Plot the vertex and the ends of the latus rectum:



Draw the latus rectum:



Sketch the parabola with that vertex and which goes through
the ends of the latus rectum:



The focus of the parabola is the midpoint of the latus
rectum, which is (-1,1), so we may as well plot the focus, too.

 

Now we know that a parabola that opens to the right 
or left has the equation:

(y - k)² = 4c(x - h)

where the vertex is (h,k)

We also know that

c = the directed distance from the vertex to the focus.
That means if we must move to the right to go from the 
vertex to the focus, then c is positive, and if we must
move to the left, then c is negative. 

the focus is the midpoint of the latus rectum. 

The length of the latus rectum is |4c|.

Now our graph tells us that c is negative because we
must go left to go from the vertex to the focus.

It also tells us that the length of the latus rectum
is 12 by simply counting the units.

We know that c = -3 for two reasons. One reason is
that it is the directed distance from the vertex to
the focus, and that is 3 units going to the left, so
c = -3.  The other reason is that the distance from
the vertex to the focus in absolute value is 1/4 of
the length of the latus rectum, or 1/4 of 12.

We are given that the vertex is (h,k) = (2,1).  So
we substitute that into

(y - k)² = 4c(x - h)

and get

(y - 1)² = 4(-3)(x - 2)

(y - 1)² = -12(x - 2)

This is the same as the second choice above.

Edwin